Meager sets of periodic functions

Jordan Bell
February 6, 2015

The following is often useful.11 1 Walter Rudin, Real and Complex Analysis, third ed., p. 68, Theorem 3.12.

Theorem 1.

If (X,μ) is a measure space, 1p, and fnLp(μ) is a sequence that converges in Lp(μ) to some fLp(μ), then there is a subsequence of fn that converges pointwise almost everywhere to f.


Assume that 1p<. For each n there is some an such that




Let ϵ>0. We have

{xX:lim supn|fan(x)-f(x)|>ϵ}N=1n=N{xX:|fan(x)-f(x)|>ϵ}.

For any N, this gives, using Chebyshev’s inequality,

μ({xX:lim supn|fan(x)-f(x)|>ϵ})n=Nμ({xX:|fan(x)-f(x)|>ϵ})ϵ-pn=Nfan-fpp.

Because n=1fan-fpp<, we have n=Nfan-fpp0 as N, which implies that

μ({xX:lim supn|fan(x)-f(x)|>ϵ})=0.

This is true for each ϵ>0, hence

μ({xX:lim supn|fan(x)-f(x)|>0})=0,

which means that for almost all xX,


Assume that p=. Let


The measure of each of these sets is 0, so for


we have μ(E)=0. For xE,


showing that for almost all xX, fk(x)f(x). ∎

The following results are in the pattern of A being a strict subset of X implying that A is meager in X.

We first work out two proofs of the following theorem.

Theorem 2.

For 1<p, Lp(T) is a meager subset of L1(T).


For n1, let


Let n1. If a sequence fkCn converges in L1(𝕋) to some fL1(𝕋), then there is a subsequence fak of fk such that for almost all x𝕋, fak(x)f(x), and so fak(x)pf(x)p. Applying the dominated convergence theorem gives


hence fpn, showing that fCn. Therefore, Cn is a closed subset of L1(𝕋) On the other hand, let fCn and let gL1(𝕋)Lp(𝕋). Then f+1kgf in L1(𝕋), and for each k we have f+1kgCn, as that would imply gLp(𝕋). This shows that f does not belong to the interior of Cn. Because Cn is closed and has empty interior, it is nowhere dense. Therefore


is meager in L1(𝕋). ∎


The open mapping theorem tells us that if X is an F-space, Y is a topological vector space, Λ:XY is continuous and linear, and Λ(X) is not meager in Y, then Λ(X)=Y, Λ is an open mapping, and Y is an F-space.22 2 Walter Rudin, Functional Analysis, second ed., p. 48, Theorem 2.11.

Let j:Lp(𝕋)L1(𝕋) be the inclusion map. For fLp(𝕋),


showing that the inclusion map is continuous. On the other hand, j is not onto, so the open mapping theorem tells us that j(Lp(𝕋))=Lp(𝕋) is meager in L1(𝕋). ∎

Suppose that X is a topological vector space, that Y is an F-space, and that Λn is a sequence of continuous linear maps XY. Let L be the set of those xX such that


exists. It is a consequence of the uniform boundedness principle that if L is not meager in X, then L=X and Λ:XY is continuous.33 3 Walter Rudin, Functional Analysis, second ed., p. 45, Theorem 2.7.

For n1, define Λn:L2(𝕋) by



L={fL2(𝕋):limnΛnf exists}.

The sequence tk=1neiktk is a Cauchy sequence in L2(𝕋), hence converges to some fL2(𝕋), which satisfies




meaning that fL2(𝕋)L. This shows that LL2(𝕋). Therefore, the above consequence of the uniform boundedness principle tells us that L is meager.