# Weak convergence and convergence in measure implies convergence in L^1?

This is a problem from S.J.Taylor Introduction to Measure and Integration p.182

Problem Let $(\Omega, \mathcal F, \mu)$ be a measure space. Let $f_n$ be a sequence of functions in $L^1(\mu)$ such that $f_n$ converge in measure to $f$ and $\left(\int_E f_n \right)$ is a Cauchy sequence for every measurable set $E$, then we must prove that $f_n$ converge to $f$ in $L^1$.

What I was trying to do is to proof that the measures $\nu_n$ induced by $f_n$ are equicontinous at $\emptyset$, this is, for all $\epsilon > 0$ and all decreasing sequence of sets $(B_k) \downarrow \emptyset$ there exists a $k_0$ such that $k\geq k_0$ implies $|\nu_n(B_k)|< \epsilon$. Because there is a result that convergence in measure and equicontinuity at $\emptyset$ implies $L^1$ convergence but the problem is that I don't know how to use effectively the condition that the measures converge for every set. One option was first consider the case where all the functions are positive or use contradiction and suppose that the sequence is not equicontinous.

• Is $\mu$ a finite measure? – saz Nov 27 '17 at 16:44
• No, Its sigma finite though – JPaucar Nov 27 '17 at 16:45
• In case it were finite I think using egoroff's theorem you can prove it, although I would have to check if my idea is correct. – JPaucar Nov 27 '17 at 16:48
• The case that $\mu$ is a finite measure is essentially discussed here: e-periodica.ch/cntmng?pid=ens-001:2001:47::182 (Note that there is a simplified version of the proof at page 277/278.) – saz Nov 27 '17 at 17:13
• The equi-continuity you are referring to is the main part of the proof what is called Vitali - Hahn - Saks Theorem. – Kavi Rama Murthy Dec 15 '17 at 7:52

The proof is the same as in cauchy-sequence-in-mean. I will only indicate the changes needed if you do not assume that $\mu$ is not finite. I use $\mathfrak{M}$ to denote the $\sigma$-algebra as in my previous answer. Instead of $\mathfrak{M}$ you consider the family $$\mathfrak{M}':=\{E\in \mathfrak{M}:\, \mu(E)<\infty\}.$$ You can identify $\mathfrak{M}'$ with a closed subset of $L^1$. Then you define $\mathcal{C}_{k}$ as before, using $\mathfrak{M}'$ instead of $\mathfrak{M}$. To prove that $\mathcal{C}_{k}$ is closed in $L^1$ you consider $\left\{ \chi_{E_{j}}\right\} \subset\mathcal{C}_{k}$ converging in $L_{1}$ to some function $f$. By extracting a subsequence (not relabeled), if necessary, you can assume that $\chi_{E_{j}}\left( x\right) \rightarrow f(x)$ for $\mu$ a.e. $x\in X$ and that $\chi_{E_{j}}(x)\le g(x)$ for all $x$ and $j$ and for some integrable function $g$ (see Lemma below). Hence, you can apply Lebesgue dominated convergence theorem as before to conclude that $\mathcal{C}_{k}$ is closed. You then continue as before and use Baire's theorem.
Lemma Assume that $f_n\to f$ in $L^1$. Then there is a subsequence such that $|f_n(x)|\le g(x)$ for some integrable function $g$.
Proof: For every $k\in\mathbb{N}$ find an integer $n_{k}\in\mathbb{N}$ such that $$\int_{X}\left\vert f_{n_{k}}\,-f\right\vert \,d\mu\leq\frac{1}{2^{k}}.$$ The sequence $\left\{ n_{k}\right\} _{k\in\mathbb{N}}$ can be chosen to be strictly increasing. For $x\in X$ define $$w\left( x\right) :=\sum_{k=1}^{\infty}\left\vert f_{n_{k}}\left( x\right) -f\left( x\right) \right\vert .$$ Then $$\int_{X}w\,d\mu\leq1.$$ Hence $w\left( x\right) <\infty$ for $\mu$ a.e. $x\in X$. At any such point $x\in X$ we have that $$\lim_{k\rightarrow\infty}\left\vert f_{n_{k}}\left( x\right) -f\left( x\right) \right\vert =0.$$ Finally, observe that $$\left\vert f_{n_{k}}\left( x\right) \right\vert \leq\left\vert f_{n_{k}% }\left( x\right) -f\left( x\right) \right\vert +\left\vert f\left( x\right) \right\vert \leq w\left( x\right) +\left\vert f\left( x\right) \right\vert =:g\left( x\right)$$ for all $x\in X$ and $k\in\mathbb{N}$. This concludes the proof.
• Very nice! Maybe this application of Baire theorem could be mentioned in this thread: math.stackexchange.com/questions/165696/…. It seems that the assumption of $\sigma$-finiteness is not needed. – Davide Giraudo Dec 16 '17 at 10:27