1
$\begingroup$

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 $\sum_k \int \vert f_k - f\vert d \mu < \infty \Rightarrow \int \vert f_k - f \vert d \mu$ $\rightarrow 0$.

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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
1
$\begingroup$

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 $X\setminus E$.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.