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 :

By contradiction

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.

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


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