$\lambda$-almost-everywhere convergence implied by lim$\lambda (${$x \in E : | f_n(x) - f(x) | > \epsilon$}$) = 0$

Question

Let $E \subset \mathbb{R}$ be $\lambda$-measurable and let $f_n,f: E -> \mathbb{R}$ $\lambda$-integrable, so that for all $\epsilon > 0$

$\lambda(${$x\in E: |f_n(x)-f(x)| > \epsilon$}$)->0$ as $n -> \infty$

Proof that $f_n$ converges to $f$ $\lambda$-almost-everywhere.

My attempt:

Assume $f_n$ does not converge almost everywhere to $f$ then there exists an $ \epsilon$ for which for all $N \in \mathbb{N}$ there's a $n>=N$ with $|f_n(x)-f(x)|>\epsilon$ for all $x \in A$ with $A$ being a set that's measure isn't zero.

Then x is in infinitely many $H_n^\epsilon:=${$x \in E : |f_n(x)-f(x)|>\epsilon$} and thus $\sum_{n=1}^\infty\lambda(H_n^\epsilon)=\infty$.

On the other hand, since $f_n,f$ are integrable we get:

$\infty>\int_E|f_n(x)-f(x)|d\lambda>\int_{H_n^\epsilon}|f_n(x)-f(x)|d\lambda=\sum_{n}\int_{H_n^\epsilon}|f_n(x)-f(x)|d\lambda>\sum_{n}\epsilon \lambda(H_n^\epsilon)=\infty$

What is a contradiction. And the assumption must be wrong.

I guess this is wrong as I didn't even use $\lambda(${$x\in E: |f_n(x)-f(x)| > \epsilon$}$)->0$ as $n -> \infty$. However, I do not find my mistake nor have I an idea on how to approach this problem.

I would really appreciate any help.

Answer 1

Fix $k$ and pick $n_k$ such that $\lambda E_k \le {1 \over 2^k}$ where $E_k =\{ x | |f_{n_k}(x)-f(x)| > { 1\over k} \}$.

This is the Borel Cantelli lemma: Note that $\int \sum_k 1_{E_k} = \sum_k \int 1_{E_k} = \sum_{k} \lambda E_k < \infty$ and so $\sum_k 1_{E_k}$ is finite ae. Hence for ae. $x$, $x$ is in at most a finite number of $E_k$ and so $f_{n_k}(x) \to f(x)$.

Addendum:

The convergence cannot be everywhere. Here is an example that is easier to describe than to write an explicit expression. Choose the space $[0,1]$, let $f = 0$ and define $f_n$ as follows: $f_1 = 1_{[0,{1 \over 2}]}, f_2 = 1_{[{1 \over 2},1]}$, $f_3 = 1_{[0,{1 \over 4}]}$, $f_4 = 1_{[{1 \over 4}, {1 \over 2}]}$, $f_5 = 1_{[{1 \over 2}, {3 \over 4}]}$, etc. It should be clear that $\lambda \{ x | |f_n(x)| > \epsilon\} \to 0$, but for every $x \in [0,1)$, we have $f_n(x) = 1$ infinitely often and $f_n(x) = 0$ and infinitely often.

However, if we pick the subsequence corresponding to the functions $1_{[0,{1 \over 2^n}]}$ then we see that this subsequence converges pointwise to $0$ for $x \in (0,1]$.