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

1 Measurable spaces

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

Theorem 1.

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

$$g_k(x)=\underset{j\ge k}{sup}{f}_j(x),h_k(x)=\underset{j\ge k}{inf}{f}_j(x),$$

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

$$g(x)=\underset{j\to \mathrm{\infty }}{lim\; sup}{f}_j(x),h(x)=\underset{j\to \mathrm{\infty }}{lim\; inf}{f}_j(x),$$

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

Proof.

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

$$\bigcup _{j=k}^{\mathrm{\infty }}{f}_j^{-1}(a,\mathrm{\infty }]\in \mathrm{\Sigma }.$$

For each $j\ge k$,

$$f_j^{-1}(a,\mathrm{\infty }]\subset \{x\in \mathrm{\Omega }:\underset{i\ge k}{sup}{f}_i(x)>a\},$$

so

$$\bigcup _{j=k}^{\mathrm{\infty }}{f}_j^{-1}(a,\mathrm{\infty }]\subset \{x\in \mathrm{\Omega }:\underset{i\ge k}{sup}{f}_i(x)>a\}.$$

If $y\notin {\bigcup }_{j=k}^{\mathrm{\infty }}{f}_j^{-1}(a,\mathrm{\infty }]$, then for each $j\ge k$, $f_j(y)\le a$, hence $g_k(y)\le a$, which means that $y\notin \{x\in \mathrm{\Omega }:g_k(x)>a\}$. Therefore,

$$\bigcup _{j=k}^{\mathrm{\infty }}{f}_j^{-1}(a,\mathrm{\infty }]=\{x\in \mathrm{\Omega }:g_k(x)>a\},$$

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

That $h_k$ is measurable follows from the fact that if $f:\mathrm{\Omega }\to \overline{ℝ}$ is measurable then $-f:\mathrm{\Omega }\to \overline{ℝ}$ is measurable, and that $h_k(x)={inf}_{j\ge k}{f}_j(x)=-{sup}_{j\ge k}(-f_j(x))$.

For $x\in \mathrm{\Omega }$,

$$g(x)=\underset{k\ge 1}{inf}{g}_k(x),$$

and because each $g_k$ is measurable it follows that $g$ is measurable. Likewise,

$$h(x)=\underset{k\ge 1}{sup}{h}_k(x),$$

and because each $h_k$ is measurable it follows that $h$ is measurable. ∎

2 Convergence in measure

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

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

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

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

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.

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 $ϵ>0$,

$$\underset{n\to \mathrm{\infty }}{lim}\mu \left(\{x\in \mathrm{\Omega }:|f_n(x)|\ge ϵ\}\right)=0.$$

Proof.

Suppose that $f_n\to 0$ in the topology of convergence in measure and let $ϵ>0$. For each $n$, let

$$A_n=\{x\in \mathrm{\Omega }:|f_n(x)|\ge ϵ\}=\{x\in \mathrm{\Omega }:\frac{|f_n(x)|}{1+|f_n(x)|}\ge \frac{ϵ}{1+ϵ}\}.$$

Because $\frac{ϵ}{1+ϵ}{\chi }_{A_n}\le \frac{|f_n|}{1+|f_n|}$,

$${\int }_{\mathrm{\Omega }}\frac{ϵ}{1+ϵ}{\chi }_{A_n}𝑑\mu \le {\int }_{\mathrm{\Omega }}\frac{|f_n|}{1+|f_n|}𝑑\mu =\rho (f_n,0),$$

i.e. $\mu (A_n)\le \frac{1+ϵ}{ϵ}\rho (f_n,0)$, which tends to $0$ as $n\to \mathrm{\infty }$.

Let $ϵ>0$ and for each $n$, let

$$A_n=\{x\in \mathrm{\Omega }:|f_n(x)|\ge ϵ\}=\{x\in \mathrm{\Omega }:\frac{|f_n(x)|}{1+|f_n(x)|}\ge \frac{ϵ}{1+ϵ}\}.$$

Suppose that $\mu (A_n)\to 0$ as $n\to \mathrm{\infty }$. There is some $n_{ϵ}$ such that $n\ge n_{ϵ}$ implies that . For $n\ge n_{ϵ}$,

$\rho (f_n,0)$ $={\displaystyle {\int }_{A_n}}{\displaystyle \frac{|f_n|}{1+|f_n|}}𝑑\mu +{\displaystyle {\int }_{\mathrm{\Omega }\setminus A_n}}{\displaystyle \frac{|f_n|}{1+|f_n|}}𝑑\mu$

$\le {\displaystyle {\int }_{A_n}}1𝑑\mu +{\displaystyle {\int }_{\mathrm{\Omega }\setminus A_n}}{\displaystyle \frac{ϵ}{1+ϵ}}𝑑\mu$

$={\displaystyle \frac{1+ϵ}{1+ϵ}}\mu (A_n)+{\displaystyle \frac{ϵ}{1+ϵ}}\mu (\mathrm{\Omega }\setminus A_n)$

$={\displaystyle \frac{1}{1+ϵ}}\mu (A_n)+{\displaystyle \frac{ϵ}{1+ϵ}}(\mu (A_n)+\mu (\mathrm{\Omega }\setminus A_n))$

$={\displaystyle \frac{1}{1+ϵ}}\mu (A_n)+{\displaystyle \frac{ϵ}{1+ϵ}}$

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 $0$.

Theorem 3.

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

Proof.

Let $ϵ>0$ and let $\eta >0$. Egorov’s theorem tells us that there is some $E\in \mathrm{\Sigma }$ with such that $f_n\to 0$ uniformly on $\mathrm{\Omega }\setminus E$. So there is some $n_0$ such that if $n\ge n_0$ and $x\in \mathrm{\Omega }\setminus E$ then . Thus for $n\ge n_0$,

$\mu (\{x\in \mathrm{\Omega }:|f_n(x)|\ge ϵ\})$ $\le \mu (E)+\mu (\{x\in \mathrm{\Omega }\setminus E:|f_n(x)|\ge ϵ\})$

$=\mu (E)$

Then $\mu (\{x\in \mathrm{\Omega }:|f_n(x)|\ge ϵ\})\to 0$ as $n\to \mathrm{\infty }$, 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 $ϵ>0$ and let $A_n=\{x\in \mathrm{\Omega }:|f_n(x)|\ge ϵ\}$. By Chebyshev’s inequality,

$$\mu (A_n)\le \frac{1}{ϵ}{\parallel f_n\parallel }_1,$$

hence $\mu (A_n)\to 0$ as $n\to \mathrm{\infty }$, 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 $0$.

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 $x\in \mathrm{\Omega }$, $f_{a(n)}(x)\to 0$.

Proof.

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

For each $n$, let

$$E_n=\{x\in \mathrm{\Omega }:|f_{a(n)}(x)|\ge \frac{1}{n}\},$$

for which . Let

$$E=\bigcap _{n=1}^{\mathrm{\infty }}\bigcup _{m=n}{E}_m,$$

and for each $n$,

Because this is true for all $n$, $\mu (E)=0$. If $x\in \mathrm{\Omega }\setminus E$, then there is some $n_x$ such that $x\notin {\bigcup }_{m=n_x}^{\mathrm{\infty }}{E}_m$. This means that for $m\ge n_x$ we have $x\notin E_m$, i.e. . This implies that for $x\notin E$, $f_{a(n)}(x)\to 0$ as $n\to \mathrm{\infty }$, showing that for almost all $x\in \mathrm{\Omega }$, $f_{a(n)}(x)\to 0$ as $n\to \mathrm{\infty }$. ∎

We now prove that $\rho$ is a complete metric, namely that $L^0(\mu )$ with this metric is an $F$-space.

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 $n,m\ge a_N$ implies that and a fortiori . Thus we presume that $f_n$ itself satisfies for $m\ge n$.

Let

$$A_{k,m}(ϵ)=\{x\in \mathrm{\Omega }:|f_k(x)-f_m(x)|\ge ϵ\}=\{x\in \mathrm{\Omega }:\frac{|f_k(x)-f_m(x)|}{1+|f_k(x)-f_m(x)|}\ge \frac{ϵ}{1+ϵ}\},$$

for which

$$\frac{ϵ}{1+ϵ}{\chi }_{A_{k,m}(ϵ)}\le \frac{|f_k(x)-f_m(x)|}{1+|f_k(x)-f_m(x)|}.$$

If $m\ge k$,

(1)

For $n=1$ and ${ϵ}_1=\frac{1}{2^1}$, let $k_1$ be such that $\frac{1+{ϵ}_1}{{ϵ}_1}\frac{1}{k_1}\le \frac{1}{2^1}$, i.e. $k_1\ge 2^1\cdot \frac{1+{ϵ}_1}{{ϵ}_1}$. For $n\ge 1$ and ${ϵ}_n=\frac{1}{2^n}$, assume that $k_n$ satisfies $\frac{1+{ϵ}_n}{{ϵ}_n}\frac{1}{k_n}\le \frac{1}{2^n}$ and $k_n>k_{n-1}$. For ${ϵ}_{n+1}=\frac{1}{2^{n+1}}$, let $k_{n+1}$ be such that $\frac{1+{ϵ}_{n+1}}{{ϵ}_{n+1}}\frac{1}{k_{n+1}}\le \frac{1}{2^{n+1}}$ and $k_{n+1}>k_n$. For any $n$ we have, because $k_n\ge n$, $\frac{1+{ϵ}_n}{{ϵ}_n}\frac{1}{k_n}\le \frac{1}{2^{k_n}}$. Then using (1) with $m\ge k_n$,

(2)

Let $g_n=f_{k_n}$ and let

$$E_n=\{x\in \mathrm{\Omega }:|g_{n+1}(x)-g_n(x)|\ge \frac{1}{2^n}\},$$

for which, by (2), . Let

$$F_n=\bigcup _{r=n}^{\mathrm{\infty }}{E}_n,$$

which satisfies

Hence $F={\bigcap }_{n=1}^{\mathrm{\infty }}{F}_n$ satisfies $\mu (F)=0$.

If $x\notin F$, then there is some $n$ for which $x\notin F_n$, i.e. for each $r\ge n$ we have $x\notin E_r$, i.e. for each $r\ge n$ we have . This implies that for $k\ge n$ and for any positive integer $p$,

$|g_{k+p}(x)-g_k(x)|$ $\le |g_{k+p}(x)-g_{k+p-1}(x)|+\mathrm{\cdots }+|g_{k+1}(x)-g_k(x)|$

so if $j\ge k$ then

(3)

This shows that if $x\notin F$ then $g_k(x)$ is a Cauchy sequence in $ℂ$, and hence converges. We define $g:\mathrm{\Omega }\to ℂ$ by

$$g(x)={\chi }_{\mathrm{\Omega }\setminus F}(x)\underset{k\to \mathrm{\infty }}{lim\; sup}{g}_k(x),x\in \mathrm{\Omega },$$

and by Theorem 1, $g\in L^0(\mu )$. For $x\notin F_k$ we have $x\notin F$ and so $g_l(x)\to g(x)$ as $l\to \mathrm{\infty }$. Then for $x\notin F_k$ and $j\ge k$, using (3) we have

as $l\to \mathrm{\infty }$, so . For $ϵ>0$, let $k$ be such that $2^{-k+1}\le ϵ$. Then

which tends to $0$ as $k\to \mathrm{\infty }$, 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 $(\mathrm{\Omega },\mathrm{\Sigma },\mu )$ be a probability space. A subset $ℱ$ of $L^1(\mu )$ is said to be equi-integrable if for every $ϵ>0$ there is some $\delta >0$ such that for all $E\in \mathrm{\Sigma }$ with $\mu (E)\le \delta$ and for all $f\in ℱ$,

$${\int }_E|f|𝑑\mu \le ϵ.$$

In other words, to say that $ℱ$ is equi-integrable means that

$$\underset{\mu (E)\to 0}{lim}\underset{f\in ℱ}{sup}{\int }_E|f|𝑑\mu =0.$$

The following theorem gives an equivalent condition for a bounded subset of $L^1(\mu )$ to be equi-integrable.

Theorem 7.

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

1. $ℱ$ is equi-integrable.

2. ${lim}_{C\to \mathrm{\infty }}{sup}_{f\in ℱ}{\int }_{\{|f|>C\}}|f|𝑑\mu =0$.

Proof.

Let and suppose that $ℱ$ is equi-integrable. For $f\in ℱ$, Chebyshev’s inequality tells us

$$\mu (\{|f|>M\})\le \frac{{\parallel f\parallel }_1}{M}\le \frac{K}{M}.$$

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

$$\underset{M\to \mathrm{\infty }}{lim}\underset{f\in ℱ}{sup}{\int }_{\{|f|>M\}}|f|𝑑\mu =0.$$

Suppose now that

$$\underset{M\to \mathrm{\infty }}{lim}\underset{f\in ℱ}{sup}{\int }_{\{|f|>M\}}|f|𝑑\mu =0.$$ (4)

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

${\displaystyle {\int }_E}|f|𝑑\mu$ $={\displaystyle {\int }_{E\cap \{|f|\le M\}}}|f|𝑑\mu +{\displaystyle {\int }_{E\cap \{|f|>M\}}}|f|𝑑\mu$

$\le M\mu (E)+{\displaystyle {\int }_{\{|f|>M\}}}|f|𝑑\mu$

$\le M\mu (E)+\underset{g\in ℱ}{sup}{\displaystyle {\int }_{\{|g|>M\}}}|g|𝑑\mu ,$

hence

$$\underset{f\in ℱ}{sup}{\int }_E|f|𝑑\mu \le M\mu (E)+\underset{g\in ℱ}{sup}{\int }_{\{|g|>M\}}|g|𝑑\mu .$$ (5)

Let $ϵ>0$. By (4) there is some $M$ such that ${sup}_{g\in ℱ}{\int }_{\{|g|>M\}}|g|𝑑\mu \le \frac{ϵ}{2}$. For $\delta =\frac{ϵ}{2M}$, if $E\in \mathrm{\Sigma }$ and $\mu (E)\le \delta$ then (5) yields

$$\underset{f\in ℱ}{sup}{\int }_E|f|𝑑\mu \le M\delta +\frac{ϵ}{2}=ϵ,$$

showing that $ℱ$ is equi-integrable. ∎

Theorem 8 (Absolute continuity of Lebesgue integral).

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

$${\int }_E|f|𝑑\mu \le ϵ.$$

Proof.

For $n\ge 1$, define

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

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

$$\underset{n\to \mathrm{\infty }}{lim}{\int }_{\mathrm{\Omega }}{g}_n𝑑\mu ={\int }_{\mathrm{\Omega }}|f|𝑑\mu .$$

Then there is some $N$ for which

$$0\le {\int }_{\mathrm{\Omega }}|f|𝑑\mu -{\int }_{\mathrm{\Omega }}{g}_N𝑑\mu \le \frac{ϵ}{2}.$$

For $\delta =\frac{ϵ}{2N}$ and $E\in \mathrm{\Sigma }$ with $\mu (E)\le \delta$,

${\displaystyle {\int }_E}|f|𝑑\mu$ $={\displaystyle {\int }_E}(|f|-g_N)𝑑\mu +{\displaystyle {\int }_E}{g}_N𝑑\mu$

$\le {\displaystyle \frac{ϵ}{2}}+{\displaystyle {\int }_E}{g}_N𝑑\mu$

$\le {\displaystyle \frac{ϵ}{2}}+N\mu (E)$

$\le {\displaystyle \frac{ϵ}{2}}+N\delta$

$=ϵ.$

∎

The following is the Vitali convergence theorem.

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

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 . Fatou’s lemma tells us that $|f|\in L^1(\mu )$ and

$${\parallel f\parallel }_1\le \underset{n\to \mathrm{\infty }}{lim\; inf}{\parallel f_{a(n)}\parallel }_1\le K.$$

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 $ϵ>0$. Because $\{f_n\}$ is equi-integrable, there is some $\delta >0$ such that for all $E\in \mathrm{\Sigma }$ with $\mu (E)\le \delta$ and for all $n$,

$${\int }_E|g_{a(n)}|𝑑\mu \le ϵ.$$

If $E\in \mathrm{\Sigma }$ with $\mu (E)\le \delta$, then ${\chi }_E{g}_{a(n)}$ converges almost everywhere to ${\chi }_{E}f$, and ${sup}_{n\ge 1}{\parallel {\chi }_E{g}_{a(n)}\parallel }_1\le ϵ$, so by Fatou’s lemma we obtain

$${\int }_E|f|𝑑\mu ={\parallel {\chi }_{E}f\parallel }_1\le \underset{n\to \mathrm{\infty }}{lim\; inf}{\parallel {\chi }_E{g}_{a(n)}\parallel }_1\le ϵ.$$

But because $g_{a(n)}$ converges almost everywhere to $f$, by Egorov’s theorem there is some $E\in \mathrm{\Sigma }$ with $\mu (E)\le \delta$ such that $g_{a(n)}\to f$ uniformly on $\mathrm{\Omega }\setminus E$, and so there is some $n_0$ such that if $n\ge n_0$ and $x\in \mathrm{\Omega }\setminus E$ then $|g_{a(n)}(x)-f(x)|\le ϵ$. Thus for $n\ge n_0$,

${\displaystyle {\int }_{\mathrm{\Omega }}}|g_{a(n)}-f|𝑑\mu$ $={\displaystyle {\int }_{\mathrm{\Omega }\setminus E}}|g_{a(n)}-f|𝑑\mu +{\displaystyle {\int }_E}|g_{a(n)}-f|𝑑\mu$

$\le \mu (\mathrm{\Omega }\setminus E)ϵ+{\displaystyle {\int }_E}|g_{a(n)}|𝑑\mu +{\displaystyle {\int }_E}|f|𝑑\mu$

$\le ϵ+ϵ+ϵ,$

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 $ϵ>0$, let $n_0$ such that $n\ge n_0$ implies that ${\parallel f_n-f\parallel }_1\le ϵ$. For each , by Theorem 8 there is some ${\delta }_{f_n}>0$ such that for $E\in \mathrm{\Sigma }$ and $\mu (E)\le {\delta }_{f_n}$,

$${\int }_E|f_n|𝑑\mu \le ϵ,$$

and likewise there is some ${\delta }_f$ such that for $E\in \mathrm{\Sigma }$ and $\mu (E)\le f$,

$${\int }_E|f|𝑑\mu \le ϵ.$$

Let $\delta >0$ be the minimum of ${\delta }_{f_1},\mathrm{\dots },{\delta }_{f_{n-1}},{\delta }_f$. Thus if $E\in \mathrm{\Sigma }$ and $\mu (E)\le \delta$, then for ,

$${\int }_E|f_n|𝑑\mu \le ϵ,$$

and for for $n\ge n_0$,

$${\int }_E|f_n|𝑑\mu \le {\int }_E|f_n-f|𝑑\mu +{\int }_E|f|𝑑\mu \le {\parallel f_n-f\parallel }_1+{\int }_E|f|𝑑\mu \le ϵ+ϵ.$$

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

The following is the de la Vallée-Poussin criterion for equi-integrability.

Theorem 10 (de la Vallée-Poussin criterion).

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

(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 $[0,\mathrm{\infty })$ satisfying (6). Let

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

$${\int }_{\{|f|\ge C\}}|f|𝑑\mu \le {\int }_{\{|f|\ge C\}}\frac{ϵ}{M}G(|f(x)|)𝑑\mu (x)\le \frac{ϵ}{M}\cdot M=ϵ.$$

Therefore by Theorem 7, $ℱ$ is bounded and equi-integrable.

Suppose that $ℱ$ is bounded and equi-integrable. For $f\in ℱ$ and $j\ge 1$, let

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

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

$$\underset{f\in ℱ}{sup}{\int }_{\{|f|>C_n\}}|f|𝑑\mu \le 2^{-n}.$$ (7)

For $f\in ℱ$ and $n\ge 1$,

${\displaystyle {\int }_{\{|f|>C_n\}}}|f|𝑑\mu$

$={\displaystyle \sum _{j=C_n}^{\mathrm{\infty }}}j({\mu }_j(f)-{\mu }_{j+1}(f))$

$={\displaystyle \sum _{j=C_n}^{\mathrm{\infty }}}{\mu }_j(f).$

Using this and (7), for $f\in ℱ$,

${\displaystyle \sum _{n=1}^{\mathrm{\infty }}}{\displaystyle \sum _{j=C_n}^{\mathrm{\infty }}}{\mu }_j(f)$ $\le {\displaystyle \sum _{n=1}^{\mathrm{\infty }}}{\displaystyle {\int }_{\{|f|>C_n\}}}|f|𝑑\mu$

$\le {\displaystyle \sum _{n=1}^{\mathrm{\infty }}}{2}^{-n}$

$=1.$

For $n\ge 0$ we define

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

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

It is apparent that $G$ is nonnegative and nondecreasing. For $t_1,t_2\in [0,\mathrm{\infty })$, $t_1\le t_2$, by the fundamental theorem of calculus,

$$G^{\prime }(t_1)(t_2-t_1)=g(t_1)(t_2-t_1)\le G(t_2)-G(t_1),$$

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

$$G(n+1)\le g(1)+g(2)+\mathrm{\cdots }+g(n+1)={\alpha }_0+{\alpha }_1+\mathrm{\cdots }+{\alpha }_n={\alpha }_1+\mathrm{\cdots }+{\alpha }_n,$$

we get

${\displaystyle {\int }_{\mathrm{\Omega }}}G(|f(x)|)𝑑\mu (x)$

$={\displaystyle \sum _{n=1}^{\mathrm{\infty }}}({\mu }_n(f)-{\mu }_{n+1}(f))G(n+1)$

$\le {\displaystyle \sum _{n=1}^{\mathrm{\infty }}}({\mu }_n(f)-{\mu }_{n+1}(f)){\displaystyle \sum _{j=1}^n}{\alpha }_j$

$={\displaystyle \sum _{n=1}^{\mathrm{\infty }}}{\mu }_n(f){\alpha }_n$

$={\displaystyle \sum _{n=1}^{\mathrm{\infty }}}{\displaystyle \sum _{j=C_n}^{\mathrm{\infty }}}{\mu }_j(f)$

$\le 1,$

showing that , which completes the proof. ∎