# $L^{0}$, convergence in measure, equi-integrability, the Vitali convergence theorem, and the de la Vallée-Poussin criterion

Jordan Bell
June 10, 2015

## 1 Measurable spaces

Let $\overline{\mathbb{R}}=\mathbb{R}\cup\{-\infty,\infty\}$, with the order topology. We assign $\mathbb{R}$ the Borel $\sigma$-algebra. It is a fact that for $E\subset\overline{\mathbb{R}}$, $E\in\mathscr{B}_{\overline{\mathbb{R}}}$ if and only if $E\setminus\{-\infty,\infty\}\in\mathscr{B}_{\mathbb{R}}$.

###### Theorem 1.

Let $(\Omega,\Sigma)$ be a measurable space. If $f_{j}$ is a sequence of measurable functions $\Omega\to\overline{\mathbb{R}}$, then for each $k$,

 $g_{k}(x)=\sup_{j\geq k}f_{j}(x),\qquad h_{k}(x)=\inf_{j\geq k}f_{j}(x),$

are measurable $\Omega\to\overline{\mathbb{R}}$, and

 $g(x)=\limsup_{j\to\infty}f_{j}(x),\qquad h(x)=\liminf_{j\to\infty}f_{j}(x),$

are measurable $\Omega\to\overline{\mathbb{R}}$.

###### Proof.

Let $a\in\mathbb{R}$. For each $j$, $f_{j}^{-1}(a,\infty]\in\Sigma$, so

 $\bigcup_{j=k}^{\infty}f_{j}^{-1}(a,\infty]\in\Sigma.$

For each $j\geq k$,

 $f_{j}^{-1}(a,\infty]\subset\left\{x\in\Omega:\sup_{i\geq k}f_{i}(x)>a\right\},$

so

 $\bigcup_{j=k}^{\infty}f_{j}^{-1}(a,\infty]\subset\left\{x\in\Omega:\sup_{i\geq k% }f_{i}(x)>a\right\}.$

If $y\not\in\bigcup_{j=k}^{\infty}f_{j}^{-1}(a,\infty]$, then for each $j\geq k$, $f_{j}(y)\leq a$, hence $g_{k}(y)\leq a$, which means that $y\not\in\left\{x\in\Omega:g_{k}(x)>a\right\}$. Therefore,

 $\bigcup_{j=k}^{\infty}f_{j}^{-1}(a,\infty]=\left\{x\in\Omega:g_{k}(x)>a\right\},$

and thus $g_{k}^{-1}(a,\infty]\in\Sigma$. Because this is true for all $a\in\mathbb{R}$ and $\mathscr{B}_{\overline{\mathbb{R}}}$ is generated by the collection $\{(a,\infty]:a\in\mathbb{R}\}$, it follows that $g_{k}:\Omega\to\overline{\mathbb{R}}$ is measurable.

That $h_{k}$ is measurable follows from the fact that if $f:\Omega\to\overline{\mathbb{R}}$ is measurable then $-f:\Omega\to\overline{\mathbb{R}}$ is measurable, and that $h_{k}(x)=\inf_{j\geq k}f_{j}(x)=-\sup_{j\geq k}(-f_{j}(x))$.

For $x\in\Omega$,

 $g(x)=\inf_{k\geq 1}g_{k}(x),$

and because each $g_{k}$ is measurable it follows that $g$ is measurable. Likewise,

 $h(x)=\sup_{k\geq 1}h_{k}(x),$

and because each $h_{k}$ is measurable it follows that $h$ is measurable. ∎

## 2 Convergence in measure

Let $(\Omega,\Sigma,\mu)$ be a probability space. Let $L^{0}(\mu)$ be the collection of equivalence classes of measurable functions $\Omega\to\mathbb{C}$, where $\mathbb{C}$ has the Borel $\sigma$-algebra, and where two functions $f$ and $g$ are equivalent when

 $\mu\{x\in\Omega:f(x)\neq g(x)\}=0.$

$L^{0}(\mu)$ is a vector space. For $f,g\in L^{0}(\mu)$ we define

 $\rho(f,g)=\int_{\Omega}\frac{|f-g|}{1+|f-g|}d\mu.$

This is a metric on $L^{0}(\mu)$, and one proves that with this metric $L^{0}(\mu)$ is a topological vector space. We call the topology induced by $\rho$ the topology of convergence in measure.11 1 Charalambos D. Aliprantis and Kim C. Border, Infinite Dimensional Analysis: A Hitchhiker’s Guide, third ed., p. 480, Lemma 13.40.

###### Theorem 2.

Suppose that $f_{n}$ is a sequence in $L^{0}(\mu)$. $f_{n}\to 0$ in the topology of convergence in measure if and only if for each $\epsilon>0$,

 $\lim_{n\to\infty}\mu\left(\{x\in\Omega:|f_{n}(x)|\geq\epsilon\}\right)=0.$
###### Proof.

Suppose that $f_{n}\to 0$ in the topology of convergence in measure and let $\epsilon>0$. For each $n$, let

 $A_{n}=\{x\in\Omega:|f_{n}(x)|\geq\epsilon\}=\left\{x\in\Omega:\frac{|f_{n}(x)|% }{1+|f_{n}(x)|}\geq\frac{\epsilon}{1+\epsilon}\right\}.$

Because $\frac{\epsilon}{1+\epsilon}\chi_{A_{n}}\leq\frac{|f_{n}|}{1+|f_{n}|}$,

 $\int_{\Omega}\frac{\epsilon}{1+\epsilon}\chi_{A_{n}}d\mu\leq\int_{\Omega}\frac% {|f_{n}|}{1+|f_{n}|}d\mu=\rho(f_{n},0),$

i.e. $\mu(A_{n})\leq\frac{1+\epsilon}{\epsilon}\rho(f_{n},0)$, which tends to $0$ as $n\to\infty$.

Let $\epsilon>0$ and for each $n$, let

 $A_{n}=\{x\in\Omega:|f_{n}(x)|\geq\epsilon\}=\left\{x\in\Omega:\frac{|f_{n}(x)|% }{1+|f_{n}(x)|}\geq\frac{\epsilon}{1+\epsilon}\right\}.$

Suppose that $\mu(A_{n})\to 0$ as $n\to\infty$. There is some $n_{\epsilon}$ such that $n\geq n_{\epsilon}$ implies that $\mu(A_{n})<\epsilon$. For $n\geq n_{\epsilon}$,

 $\displaystyle\rho(f_{n},0)$ $\displaystyle=\int_{A_{n}}\frac{|f_{n}|}{1+|f_{n}|}d\mu+\int_{\Omega\setminus A% _{n}}\frac{|f_{n}|}{1+|f_{n}|}d\mu$ $\displaystyle\leq\int_{A_{n}}1\;d\mu+\int_{\Omega\setminus A_{n}}\frac{% \epsilon}{1+\epsilon}d\mu$ $\displaystyle=\frac{1+\epsilon}{1+\epsilon}\mu(A_{n})+\frac{\epsilon}{1+% \epsilon}\mu(\Omega\setminus A_{n})$ $\displaystyle=\frac{1}{1+\epsilon}\mu(A_{n})+\frac{\epsilon}{1+\epsilon}(\mu(A% _{n})+\mu(\Omega\setminus A_{n}))$ $\displaystyle=\frac{1}{1+\epsilon}\mu(A_{n})+\frac{\epsilon}{1+\epsilon}$ $\displaystyle<\frac{1}{1+\epsilon}\cdot\epsilon+\frac{\epsilon}{1+\epsilon}$ $\displaystyle<2\epsilon.$

This shows that $f_{n}\to 0$ in the topology of convergence in measure. ∎

We now prove that if a sequence in $L^{0}(\mu)$ converges almost everywhere to $0$ then it converges in measure to $0$.22 2 Charalambos D. Aliprantis and Kim C. Border, Infinite Dimensional Analysis: A Hitchhiker’s Guide, third ed., p. 479, Theorem 13.37.

###### Theorem 3.

Suppose that $f_{n}$ is a sequence in $L^{0}(\mu)$ and that for almost all $x\in\Omega$, $f_{n}(x)\to 0$ as $n\to\infty$. Then $f_{n}\to 0$ in the topology of convergence in measure.

###### Proof.

Let $\epsilon>0$ and let $\eta>0$. Egorov’s theorem tells us that there is some $E\in\Sigma$ with $\mu(E)<\eta$ such that $f_{n}\to 0$ uniformly on $\Omega\setminus E$. So there is some $n_{0}$ such that if $n\geq n_{0}$ and $x\in\Omega\setminus E$ then $|f_{n}(x)|<\epsilon$. Thus for $n\geq n_{0}$,

 $\displaystyle\mu(\{x\in\Omega:|f_{n}(x)|\geq\epsilon\})$ $\displaystyle\leq\mu(E)+\mu(\{x\in\Omega\setminus E:|f_{n}(x)|\geq\epsilon\})$ $\displaystyle=\mu(E)$ $\displaystyle<\eta.$

Then $\mu(\{x\in\Omega:|f_{n}(x)|\geq\epsilon\})\to 0$ as $n\to\infty$, namely, $f_{n}\to 0$ in measure. ∎

###### Theorem 4.

If $f_{n}$ is a sequence in $L^{1}(\mu)$ that converges in $L^{1}(\mu)$ to $0$, then $f_{n}$ converges in measure to $0$.

###### Proof.

Let $\epsilon>0$ and let $A_{n}=\{x\in\Omega:|f_{n}(x)|\geq\epsilon\}$. By Chebyshev’s inequality,

 $\mu(A_{n})\leq\frac{1}{\epsilon}\left\|f_{n}\right\|_{1},$

hence $\mu(A_{n})\to 0$ as $n\to\infty$, namely, $f_{n}$ converges to $0$ in measure. ∎

The following theorem shows that a sequence in $L^{0}(\mu)$ that converges in measure to $0$ then it has a subsequence that almost everywhere converges to $0$.33 3 Charalambos D. Aliprantis and Kim C. Border, Infinite Dimensional Analysis: A Hitchhiker’s Guide, third ed., p. 479, Theorem 13.38.

###### Theorem 5.

Suppose that $f_{n}$ is a sequence in $L^{0}(\mu)$ that converges in measure to $0$. Then there is a subsequence $f_{a(n)}$ of $f_{n}$ such that for almost all $x\in\Omega$, $f_{a(n)}(x)\to 0$.

###### Proof.

For each $n$, with $\epsilon=\frac{1}{n}$, there is some $a(n)$ such that $m\geq a(n)$ implies that

 $\mu\left(\left\{x\in\Omega:|f_{m}(x)|\geq\frac{1}{n}\right\}\right)<\frac{1}{2% ^{n}}.$

For each $n$, let

 $E_{n}=\left\{x\in\Omega:|f_{a(n)}(x)|\geq\frac{1}{n}\right\},$

for which $\mu(E_{n})<\frac{1}{2^{n}}$. Let

 $E=\bigcap_{n=1}^{\infty}\bigcup_{m=n}E_{m},$

and for each $n$,

 $\mu(E)\leq\mu\left(\bigcup_{m=n}^{\infty}E_{m}\right)\leq\sum_{m=n}^{\infty}% \mu(E_{m})<\sum_{m=n}^{\infty}\frac{1}{2^{m}}=\frac{1}{2^{n}}\cdot 2=2^{1-n}.$

Because this is true for all $n$, $\mu(E)=0$. If $x\in\Omega\setminus E$, then there is some $n_{x}$ such that $x\not\in\bigcup_{m=n_{x}}^{\infty}E_{m}$. This means that for $m\geq n_{x}$ we have $x\not\in E_{m}$, i.e. $|f_{a(m)}(x)|<\frac{1}{m}$. This implies that for $x\not\in E$, $f_{a(n)}(x)\to 0$ as $n\to\infty$, showing that for almost all $x\in\Omega$, $f_{a(n)}(x)\to 0$ as $n\to\infty$. ∎

We now prove that $\rho$ is a complete metric, namely that $L^{0}(\mu)$ with this metric is an $F$-space.44 4 Charalambos D. Aliprantis and Kim C. Border, Infinite Dimensional Analysis: A Hitchhiker’s Guide, third ed., p. 481, Theorem 13.41; Gerald B. Folland, Real Analysis: Modern Techniques and Their Applications, second ed., p. 61, Theorem 2.30.

###### Theorem 6.

$\rho$ is a complete metric on $L^{0}(\mu)$.

###### Proof.

Suppose that $f_{n}$ is a Cauchy sequence in $L^{0}(\mu)$. To prove that $f_{n}$ is convergent it suffices to prove that $f_{n}$ has a convergent subsequence. If $(X,d)$ is a metric space and $x_{n}$ is a Cauchy sequence in $X$, for any $N$ let $a_{N}$ be such that $n,m\geq a_{N}$ implies that $d(x_{n},x_{m})<\frac{1}{N}$ and a fortiori $d(x_{n},x_{m})<\frac{1}{a_{N}}$. Thus we presume that $f_{n}$ itself satisfies $\rho(f_{n},f_{m})<\frac{1}{n}$ for $m\geq n$.

Let

 $A_{k,m}(\epsilon)=\{x\in\Omega:|f_{k}(x)-f_{m}(x)|\geq\epsilon\}=\left\{x\in% \Omega:\frac{|f_{k}(x)-f_{m}(x)|}{1+|f_{k}(x)-f_{m}(x)|}\geq\frac{\epsilon}{1+% \epsilon}\right\},$

for which

 $\frac{\epsilon}{1+\epsilon}\chi_{A_{k,m}(\epsilon)}\leq\frac{|f_{k}(x)-f_{m}(x% )|}{1+|f_{k}(x)-f_{m}(x)|}.$

If $m\geq k$,

 $\mu(A_{k,m}(\epsilon))\leq\frac{1+\epsilon}{\epsilon}\int_{\Omega}\frac{|f_{k}% -f_{m}|}{1+|f_{k}-f_{m}|}d\mu=\frac{1+\epsilon}{\epsilon}\rho(f_{k},f_{m})<% \frac{1+\epsilon}{\epsilon}\frac{1}{k}.$ (1)

For $n=1$ and $\epsilon_{1}=\frac{1}{2^{1}}$, let $k_{1}$ be such that $\frac{1+\epsilon_{1}}{\epsilon_{1}}\frac{1}{k_{1}}\leq\frac{1}{2^{1}}$, i.e. $k_{1}\geq 2^{1}\cdot\frac{1+\epsilon_{1}}{\epsilon_{1}}$. For $n\geq 1$ and $\epsilon_{n}=\frac{1}{2^{n}}$, assume that $k_{n}$ satisfies $\frac{1+\epsilon_{n}}{\epsilon_{n}}\frac{1}{k_{n}}\leq\frac{1}{2^{n}}$ and $k_{n}>k_{n-1}$. For $\epsilon_{n+1}=\frac{1}{2^{n+1}}$, let $k_{n+1}$ be such that $\frac{1+\epsilon_{n+1}}{\epsilon_{n+1}}\frac{1}{k_{n+1}}\leq\frac{1}{2^{n+1}}$ and $k_{n+1}>k_{n}$. For any $n$ we have, because $k_{n}\geq n$, $\frac{1+\epsilon_{n}}{\epsilon_{n}}\frac{1}{k_{n}}\leq\frac{1}{2^{k_{n}}}$. Then using (1) with $m\geq k_{n}$,

 $\mu\left(A_{k_{n},m}\left(\frac{1}{2^{n}}\right)\right)<\frac{1+\epsilon_{n}}{% \epsilon_{n}}\frac{1}{k_{n}}\leq\frac{1}{2^{k_{n}}}.$ (2)

Let $g_{n}=f_{k_{n}}$ and let

 $E_{n}=\left\{x\in\Omega:|g_{n+1}(x)-g_{n}(x)|\geq\frac{1}{2^{n}}\right\},$

for which, by (2), $\mu(E_{n})<\frac{1}{2^{n}}$. Let

 $F_{n}=\bigcup_{r=n}^{\infty}E_{n},$

which satisfies

 $\mu(F_{n})\leq\sum_{r=n}^{\infty}\mu(E_{n})<\sum_{r=n}^{\infty}2^{-r}=2^{-n+1}.$

Hence $F=\bigcap_{n=1}^{\infty}F_{n}$ satisfies $\mu(F)=0$.

If $x\not\in F$, then there is some $n$ for which $x\not\in F_{n}$, i.e. for each $r\geq n$ we have $x\not\in E_{r}$, i.e. for each $r\geq n$ we have $|g_{r+1}(x)-g_{r}(x)|<2^{-r}$. This implies that for $k\geq n$ and for any positive integer $p$,

 $\displaystyle|g_{k+p}(x)-g_{k}(x)|$ $\displaystyle\leq|g_{k+p}(x)-g_{k+p-1}(x)|+\cdots+|g_{k+1}(x)-g_{k}(x)|$ $\displaystyle<2^{-k-p+1}+\cdots+2^{-k}$ $\displaystyle<2^{-k+1},$

so if $j\geq k$ then

 $|g_{j}(x)-g_{k}(x)|<2^{-k+1}.$ (3)

This shows that if $x\not\in F$ then $g_{k}(x)$ is a Cauchy sequence in $\mathbb{C}$, and hence converges. We define $g:\Omega\to\mathbb{C}$ by

 $g(x)=\chi_{\Omega\setminus F}(x)\limsup_{k\to\infty}g_{k}(x),\qquad x\in\Omega,$

and by Theorem 1, $g\in L^{0}(\mu)$. For $x\not\in F_{k}$ we have $x\not\in F$ and so $g_{l}(x)\to g(x)$ as $l\to\infty$. Then for $x\not\in F_{k}$ and $j\geq k$, using (3) we have

 $|g_{j}(x)-g(x)|\leq|g_{j}(x)-g_{l}(x)|+|g_{l}(x)-g(x)|<2^{-k+1}+|g_{l}(x)-g(x)% |\to 2^{-k+1}$

as $l\to\infty$, so $|g_{j}(x)-g(x)|<2^{-k+1}$. For $\epsilon>0$, let $k$ be such that $2^{-k+1}\leq\epsilon$. Then

 $\begin{split}&\displaystyle\mu(\{x\in\Omega:|g_{k}(x)-g(x)|\geq\epsilon\})\\ \displaystyle\leq&\displaystyle\mu(F_{k})+\mu(\{x\in\Omega\setminus F_{k}:|g_{% k}(x)-g(x)|\geq\epsilon\})\\ \displaystyle<&\displaystyle 2^{-k+1}+\mu(\{x\in\Omega\setminus F_{k}:|g_{k}(x% )-g(x)|\geq 2^{-k+1}\})\\ \displaystyle=&\displaystyle 2^{-k+1},\end{split}$

which tends to $0$ as $k\to\infty$, showing that $g_{k}$ converges to $g$ in measure, and because $g_{k}$ is a subsequence of $f_{k}$ this completes the proof. ∎

## 3 Equi-integrability

Let $(\Omega,\Sigma,\mu)$ be a probability space. A subset $\mathscr{F}$ of $L^{1}(\mu)$ is said to be equi-integrable if for every $\epsilon>0$ there is some $\delta>0$ such that for all $E\in\Sigma$ with $\mu(E)\leq\delta$ and for all $f\in\mathscr{F}$,

 $\int_{E}|f|d\mu\leq\epsilon.$

In other words, to say that $\mathscr{F}$ is equi-integrable means that

 $\lim_{\mu(E)\to 0}\sup_{f\in\mathscr{F}}\int_{E}|f|d\mu=0.$

The following theorem gives an equivalent condition for a bounded subset of $L^{1}(\mu)$ to be equi-integrable.55 5 Fernando Albiac and Nigel J. Kalton, Topics in Banach Space Theory, p. 105, Lemma 5.2.6.

###### Theorem 7.

Suppose that $\mathscr{F}$ is a bounded subset of $L^{1}(\mu)$. Then the following are equivalent:

1. 1.

$\mathscr{F}$ is equi-integrable.

2. 2.

$\lim_{C\to\infty}\sup_{f\in\mathscr{F}}\int_{\{|f|>C\}}|f|d\mu=0$.

###### Proof.

Let $K=\sup_{f\in\mathscr{F}}\left\|f\right\|_{1}<\infty$ and suppose that $\mathscr{F}$ is equi-integrable. For $f\in\mathscr{F}$, Chebyshev’s inequality tells us

 $\mu(\{|f|>M\})\leq\frac{\left\|f\right\|_{1}}{M}\leq\frac{K}{M}.$

Because $\mu(\{|f|>M\})\to 0$ as $M\to\infty$ and $\mathscr{F}$ is equi-integrable,

 $\lim_{M\to\infty}\sup_{f\in\mathscr{F}}\int_{\{|f|>M\}}|f|d\mu=0.$

Suppose now that

 $\lim_{M\to\infty}\sup_{f\in\mathscr{F}}\int_{\{|f|>M\}}|f|d\mu=0.$ (4)

For $E\in\Sigma$ and $f\in\mathscr{F}$, if $M>0$ then

 $\displaystyle\int_{E}|f|d\mu$ $\displaystyle=\int_{E\cap\{|f|\leq M\}}|f|d\mu+\int_{E\cap\{|f|>M\}}|f|d\mu$ $\displaystyle\leq M\mu(E)+\int_{\{|f|>M\}}|f|d\mu$ $\displaystyle\leq M\mu(E)+\sup_{g\in\mathscr{F}}\int_{\{|g|>M\}}|g|d\mu,$

hence

 $\sup_{f\in\mathscr{F}}\int_{E}|f|d\mu\leq M\mu(E)+\sup_{g\in\mathscr{F}}\int_{% \{|g|>M\}}|g|d\mu.$ (5)

Let $\epsilon>0$. By (4) there is some $M$ such that $\sup_{g\in\mathscr{F}}\int_{\{|g|>M\}}|g|d\mu\leq\frac{\epsilon}{2}$. For $\delta=\frac{\epsilon}{2M}$, if $E\in\Sigma$ and $\mu(E)\leq\delta$ then (5) yields

 $\sup_{f\in\mathscr{F}}\int_{E}|f|d\mu\leq M\delta+\frac{\epsilon}{2}=\epsilon,$

showing that $\mathscr{F}$ is equi-integrable. ∎

###### Theorem 8 (Absolute continuity of Lebesgue integral).

Suppose that $f\in L^{1}(\mu)$. If $\epsilon>0$ then there is some $\delta>0$ such that for any $E\in\Sigma$ with $\mu(E)\leq\delta$,

 $\int_{E}|f|d\mu\leq\epsilon.$
###### Proof.

For $n\geq 1$, define

 $g_{n}(x)=\min\{|f(x)|,n\},\qquad x\in\Omega.$

Then $g_{n}$ is a sequence in $L^{1}(\mu)$ such that for each $x\in\Omega$, $g_{n}(x)$ is nondecreasing and $g_{n}(x)\to|f(x)|$, and thus the monotone convergence theorem tells us that

 $\lim_{n\to\infty}\int_{\Omega}g_{n}d\mu=\int_{\Omega}|f|d\mu.$

Then there is some $N$ for which

 $0\leq\int_{\Omega}|f|d\mu-\int_{\Omega}g_{N}d\mu\leq\frac{\epsilon}{2}.$

For $\delta=\frac{\epsilon}{2N}$ and $E\in\Sigma$ with $\mu(E)\leq\delta$,

 $\displaystyle\int_{E}|f|d\mu$ $\displaystyle=\int_{E}(|f|-g_{N})d\mu+\int_{E}g_{N}d\mu$ $\displaystyle\leq\frac{\epsilon}{2}+\int_{E}g_{N}d\mu$ $\displaystyle\leq\frac{\epsilon}{2}+N\mu(E)$ $\displaystyle\leq\frac{\epsilon}{2}+N\delta$ $\displaystyle=\epsilon.$

The following is the Vitali convergence theorem.66 6 V. I. Bogachev, Measure Theory, volume I, p. 268, Theorem 4.5.4.

###### Theorem 9 (Vitali convergence theorem).

Suppose that $f\in L^{0}(\mu)$ and $f_{n}$ is a sequence in $L^{1}(\mu)$. Then the following are equivalent:

1. 1.

$\{f_{n}\}$ is equi-integrable, $\{f_{n}\}$ is bounded in $L^{1}(\mu)$, and $f_{n}\to f$ in measure.

2. 2.

$f\in L^{1}(\mu)$ and $f_{n}\to f$ in $L^{1}(\mu)$.

###### Proof.

Suppose that $\{f_{n}\}$ is equi-integrable and $f_{n}\to f$ in measure. Because $f_{n}\to f$ in measure, there is a subsequence $f_{a(n)}$ of $f_{n}$ that converges almost everywhere to $f$ and so $|f_{a(n)}|$ converges almost everywhere to $|f|$. Let $K=\sup_{n\geq 1}\left\|f_{a(n)}\right\|_{1}<\infty$. Fatou’s lemma tells us that $|f|\in L^{1}(\mu)$ and

 $\left\|f\right\|_{1}\leq\liminf_{n\to\infty}\left\|f_{a(n)}\right\|_{1}\leq K.$

Because $f\in L^{0}(\mu)$ and $|f|\in L^{1}(\mu)$, $f\in L^{1}(\mu)$. To show that $f_{n}$ converges to $f$ in $L^{1}(\mu)$, it suffices to show that any subsequence of $f_{n}$ itself has a subsequence that converges to $f$ in $L^{1}(\mu)$. (Generally, a sequence in a topological space converges to $x$ if and only if any subsequence itself has a subsequence that converges to $x$.) Thus, let $g_{n}$ be a subsequence of $f_{n}$. Because $f_{n}$ converges to $f$ in measure, the subsequence $g_{n}$ converges to $f$ in measure and so there is a subsequence $g_{a(n)}$ of $g_{n}$ that converges almost everywhere to $f$. Let $\epsilon>0$. Because $\{f_{n}\}$ is equi-integrable, there is some $\delta>0$ such that for all $E\in\Sigma$ with $\mu(E)\leq\delta$ and for all $n$,

 $\int_{E}|g_{a(n)}|d\mu\leq\epsilon.$

If $E\in\Sigma$ with $\mu(E)\leq\delta$, then $\chi_{E}g_{a(n)}$ converges almost everywhere to $\chi_{E}f$, and $\sup_{n\geq 1}\left\|\chi_{E}g_{a(n)}\right\|_{1}\leq\epsilon$, so by Fatou’s lemma we obtain

 $\int_{E}|f|d\mu=\left\|\chi_{E}f\right\|_{1}\leq\liminf_{n\to\infty}\left\|% \chi_{E}g_{a(n)}\right\|_{1}\leq\epsilon.$

But because $g_{a(n)}$ converges almost everywhere to $f$, by Egorov’s theorem there is some $E\in\Sigma$ with $\mu(E)\leq\delta$ such that $g_{a(n)}\to f$ uniformly on $\Omega\setminus E$, and so there is some $n_{0}$ such that if $n\geq n_{0}$ and $x\in\Omega\setminus E$ then $|g_{a(n)}(x)-f(x)|\leq\epsilon$. Thus for $n\geq n_{0}$,

 $\displaystyle\int_{\Omega}|g_{a(n)}-f|d\mu$ $\displaystyle=\int_{\Omega\setminus E}|g_{a(n)}-f|d\mu+\int_{E}|g_{a(n)}-f|d\mu$ $\displaystyle\leq\mu(\Omega\setminus E)\epsilon+\int_{E}|g_{a(n)}|d\mu+\int_{E% }|f|d\mu$ $\displaystyle\leq\epsilon+\epsilon+\epsilon,$

which shows that $g_{a(n)}\to f$ in $L^{1}(\mu)$. That is, we have shown that for any subsequence $g_{n}$ of $f_{n}$ there is a subsequence $g_{a(n)}$ of $g_{n}$ that converges to $f$ in $L^{1}(\mu)$, which implies that the sequence $f_{n}$ converges to $f$ in $L^{1}(\mu)$.

Suppose that $f\in L^{1}(\mu)$ and $f_{n}\to f$ in $L^{1}(\mu)$. First, because the sequence $f_{n}$ is convergent in $L^{1}(\mu)$ the set $\{f_{n}\}$ is bounded in $L^{1}(\mu)$. Second, $f_{n}\to f$ in $L^{1}(\mu)$ implies that $f_{n}\to f$ in measure. Third, for $\epsilon>0$, let $n_{0}$ such that $n\geq n_{0}$ implies that $\left\|f_{n}-f\right\|_{1}\leq\epsilon$. For each $1\leq n, by Theorem 8 there is some $\delta_{f_{n}}>0$ such that for $E\in\Sigma$ and $\mu(E)\leq\delta_{f_{n}}$,

 $\int_{E}|f_{n}|d\mu\leq\epsilon,$

and likewise there is some $\delta_{f}$ such that for $E\in\Sigma$ and $\mu(E)\leq f$,

 $\int_{E}|f|d\mu\leq\epsilon.$

Let $\delta>0$ be the minimum of $\delta_{f_{1}},\ldots,\delta_{f_{n-1}},\delta_{f}$. Thus if $E\in\Sigma$ and $\mu(E)\leq\delta$, then for $1\leq n,

 $\int_{E}|f_{n}|d\mu\leq\epsilon,$

and for for $n\geq n_{0}$,

 $\int_{E}|f_{n}|d\mu\leq\int_{E}|f_{n}-f|d\mu+\int_{E}|f|d\mu\leq\left\|f_{n}-f% \right\|_{1}+\int_{E}|f|d\mu\leq\epsilon+\epsilon.$

This shows that $\{f_{n}\}$ is equi-integrable, completing the proof. ∎

The following is the de la Vallée-Poussin criterion for equi-integrability.77 7 V. I. Bogachev, Measure Theory, volume I, p. 272, Theorem 4.5.9.

###### Theorem 10 (de la Vallée-Poussin criterion).

Suppose that $\mathscr{F}\subset L^{1}(\mu)$. $\mathscr{F}$ is bounded and equi-integrable if and only if there is a there nonnegative nondecreasing function $G$ on $[0,\infty)$ such that

 $\lim_{t\to\infty}\frac{G(t)}{t}=\infty\quad\textrm{and}\quad\sup_{f\in\mathscr% {F}}\int_{\Omega}G(|f(x)|)d\mu(x)<\infty,$ (6)

and if there is a nonnegative nondecreasing function $G$ satisfying (6) then there is a convex nonnegative nondecreasing function $G$ satisfying (6).

###### Proof.

Suppose that $G$ is a nonnegative nondecreasing function on $[0,\infty)$ satisfying (6). Let

 $\sup_{f\in\mathscr{F}}\int_{\Omega}G(|f(x)|)d\mu(x)\leq M<\infty.$

For $\epsilon>0$, there is some $C$ such that $t\geq C$ implies that $\frac{G(t)}{t}\geq\frac{M}{\epsilon}$, and hence, for $f\in\mathscr{F}$, if $x\in\Omega$ and $|f(x)|\geq C$ then $\frac{G(|f(x)|)}{|f(x)|}\geq\frac{M}{\epsilon}$, i.e. $|f(x)|\leq\frac{\epsilon}{M}G(|f(x)|)$, which yields

 $\int_{\{|f|\geq C\}}|f|d\mu\leq\int_{\{|f|\geq C\}}\frac{\epsilon}{M}G(|f(x)|)% d\mu(x)\leq\frac{\epsilon}{M}\cdot M=\epsilon.$

Therefore by Theorem 7, $\mathscr{F}$ is bounded and equi-integrable.

Suppose that $\mathscr{F}$ is bounded and equi-integrable. For $f\in\mathscr{F}$ and $j\geq 1$, let

 $\mu_{j}(f)=\mu(\{x\in\Omega:|f(x)|>j\})\in\Sigma.$

By induction, because $\mathscr{F}$ is bounded and equi-integrable there is a strictly increasing sequence of positive integers $C_{n}$ such that for each $n$,

 $\sup_{f\in\mathscr{F}}\int_{\{|f|>C_{n}\}}|f|d\mu\leq 2^{-n}.$ (7)

For $f\in\mathscr{F}$ and $n\geq 1$,

 $\displaystyle\int_{\{|f|>C_{n}\}}|f|d\mu$ $\displaystyle=\sum_{j=C_{n}}^{\infty}\int_{\{j<|f|\leq j+1\}}|f|d\mu$ $\displaystyle\geq\sum_{j=C_{n}}^{\infty}j\mu(\{x\in\Omega:j<|f(x)|\leq j+1\})$ $\displaystyle=\sum_{j=C_{n}}^{\infty}j(\mu_{j}(f)-\mu_{j+1}(f))$ $\displaystyle=\sum_{j=C_{n}}^{\infty}\mu_{j}(f).$

Using this and (7), for $f\in\mathscr{F}$,

 $\displaystyle\sum_{n=1}^{\infty}\sum_{j=C_{n}}^{\infty}\mu_{j}(f)$ $\displaystyle\leq\sum_{n=1}^{\infty}\int_{\{|f|>C_{n}\}}|f|d\mu$ $\displaystyle\leq\sum_{n=1}^{\infty}2^{-n}$ $\displaystyle=1.$

For $n\geq 0$ we define

 $\alpha_{n}=\begin{cases}0&n

It is straightforward that $\alpha_{n}\to\infty$ as $n\to\infty$. We define a step function $g$ on $[0,\infty)$ by

 $g(t)=\sum_{n=0}^{\infty}\alpha_{n}\chi_{(n,n+1]}(t),\qquad 0\leq t<\infty,$

and we define a function $G$ on $[0,\infty)$ by

 $G(t)=\int_{0}^{t}g(s)ds,\qquad 0\leq t<\infty.$

It is apparent that $G$ is nonnegative and nondecreasing. For $t_{1},t_{2}\in[0,\infty)$, $t_{1}\leq t_{2}$, by the fundamental theorem of calculus,

 $G^{\prime}(t_{1})(t_{2}-t_{1})=g(t_{1})(t_{2}-t_{1})\leq G(t_{2})-G(t_{1}),$

showing that $G$ is convex. The above inequality also yields that for $t>0$, $\frac{G(t)}{t}\geq\frac{g(t/2)}{2}$, and $g(t/2)\to\infty$ as $t\to\infty$ so we get that $\lim_{t\to\infty}\frac{G(t)}{t}=\infty$. For $f\in\mathscr{F}$, using $G(0)=0$, $G(1)=0$, and for $n\geq 1$,

 $G(n+1)\leq g(1)+g(2)+\cdots+g(n+1)=\alpha_{0}+\alpha_{1}+\cdots+\alpha_{n}=% \alpha_{1}+\cdots+\alpha_{n},$

we get

 $\displaystyle\int_{\Omega}G(|f(x)|)d\mu(x)$ $\displaystyle=\int_{\{|f|=0\}}G(|f(x)|)d\mu(x)+\sum_{n=0}^{\infty}\int_{\{n<|f% |\leq n+1\}}G(|f(x)|)d\mu(x)$ $\displaystyle\leq\sum_{n=0}^{\infty}\int_{\{n<|f|\leq n+1\}}G(n+1)d\mu(x)$ $\displaystyle=\sum_{n=1}^{\infty}(\mu_{n}(f)-\mu_{n+1}(f))G(n+1)$ $\displaystyle\leq\sum_{n=1}^{\infty}(\mu_{n}(f)-\mu_{n+1}(f))\sum_{j=1}^{n}% \alpha_{j}$ $\displaystyle=\sum_{n=1}^{\infty}\mu_{n}(f)\alpha_{n}$ $\displaystyle=\sum_{n=1}^{\infty}\sum_{j=C_{n}}^{\infty}\mu_{j}(f)$ $\displaystyle\leq 1,$

showing that $\sup_{f\in\mathscr{F}}\int_{\Omega}G(|f(x)|)d\mu(x)<\infty$, which completes the proof. ∎