# Convergence in measure and almost everywhere proof

I'm trying to prove the following :

Let $$(X,\mathcal A,\mu)$$ be a measure space. Let $$f_n \in L^1$$ converge in measure to $$f \in L^1$$ and that

$$\sum_k \int \vert f_k - f\vert d \mu < \infty$$

Then $$f_n$$ converges almost everywhere to $$f$$.

Attempt & ideas :

Suppose $$f_n$$ doesn't converge almost everywhere to $$f$$. Then $$\not \exists A \in \mathcal A \; s.t. \; \mu(A) = 0$$ and

$$N = \{ x : f_k \not \rightarrow f \} \subset A.$$

Let x $$\in N,\exists \epsilon_x > 0,$$ s.t. $$\forall k$$ :

$$\vert f_k(x) - f(x) \vert > \epsilon_x.$$

Now the rough idea of rest of my "proof" is the following :

take $$\epsilon$$ the smallest of all $$\epsilon_x$$ then if $$\epsilon \neq 0$$ then

$$0 < \mu\{x :\vert f_k(x) - f(x)\vert > \epsilon \} < \frac{1}{\epsilon} \int \vert f_k - f \vert d \mu$$ and get a contradiction since $$\sum_k \int \vert f_k - f\vert d \mu < \infty \Rightarrow \int \vert f_k - f \vert d \mu$$ $$\rightarrow 0$$.

Fix $$\varepsilon>0$$. Note that $$\sum_{n=1}^\infty\mu(|f_n-f|>\varepsilon)\leq\sum_{n=1}^\infty\frac{1}{\varepsilon}\int|f_n-f|\, d\mu<\infty$$ by Markov's inequality. It follows by Borel Cantelli that $$\mu(|f_n-f|>\varepsilon\, \text{i.o})=0.\tag{0}$$ Note that the set $$A$$ on which the sequence $$(f_n)$$ does not converge can be written as $$A=\bigcup_{\varepsilon>0}\bigcap_{N\geq1}\bigcup_{n\geq N}(|f_n-f|>\varepsilon).\tag{1}$$ where we take $$\varepsilon$$ through a countable sequence decreasing towards zero in the union above (like $$1/n$$). A union bound applied to $$(1)$$ together with $$(0)$$ yield that $$\mu(A)=0$$ as desired.

• what does the i.o. stand for in $\mu (\vert f_n -f \vert > \epsilon$ i.o.$) = 0.$ ? – Digitalis Jan 12 '19 at 23:30
• It means "infinitely often". The set $\{A_n$ i.o$\}$ is the set of elements that belong to infinitely many of the sets $A_n$. – Mark Jan 12 '19 at 23:33

Alright, but what if $$\epsilon=0$$? Usually it is hard to take a smallest element in an infinite set.

Here is what you can do. Let any $$n\in\mathbb{N}$$. For each $$k\in\mathbb{N}$$ we have $$\mu\{x: |f_k(x)-f(x)|\geq\frac{1}{n}\}\leq n\int|f_k-f|d\mu$$. Hence using the fact that the sum of integrals converges we get $$\sum_{k=1}^\infty \mu\{x: |f_k(x)-f(x)|\geq\frac{1}{n}\}<\infty$$. And now from the Borel-Cantelli lemma we know that there exists a null set $$E_n$$ such that if $$x\in X\setminus E_n$$ then $$x$$ belongs only to a finite number of the sets $$(\{x: |f_k(x)-f(x)|\geq\frac{1}{n}\})_{k=1}^\infty$$.

Now do that for every $$n\in\mathbb{N}$$ and then define $$E=\cup_{n=1}^\infty E_n$$. It is a null set as a countable union of null sets. Now show that there is pointwise convergence in $$X\setminus E$$.

Using the dominated convergence theorem

$$\sum_k \int \vert f_k - f\vert d \mu = \int \sum_k \vert f_k - f \vert d\mu < \infty.$$

Therefore $$\sum_k \vert f_k - f \vert$$ is integrable and is finite almost everywhere and

$$\lim_{k \to \infty} \vert f_k - f \vert = 0 \qquad \mu.a.e.$$

and the sequence $$f_k$$ converges to $$f$$ almost everywhere.