Converges in measure then a subsequence converges almost everywhere

Let $(X,\mathcal{M},\mu)$ be a measure space. Let $\{f_n\}_{n\in\mathbb{N}}$ a sequence of measurable functions which converge in measure to a function $f$. Prove that exists a subsequence $\{f_{n_k}\}_{k\in\mathbb{N}}$ such that converges almost everywhere to $f$.

Attempt: (Since I not used the Borel-Cantelli Lemma I'm not sure if it's right, I would appreciate a lot if you check)

Since $f_n$ converges in measure to $f$ we have the following:

$$\forall\epsilon>0\ \ \forall\delta>0\ \ \exists\, n_o\in\mathbb{N}\, \big(n\geq n_o \Rightarrow \mu(x\in X : |f(x)-f_n(x)|\geq\epsilon)<\delta\big)$$

For each $m\in\mathbb{N}$ we define $\epsilon_m=\frac{1}{m}$ and the following subsequence of $f_n$. Taking $\delta_n=\frac{1}{n}$ exist $n_{m,n}\in\mathbb{N}$ such that $\mu(x\in X : |f(x)-f_{\displaystyle n_{m,n}}(x)|\geq\frac{1}{m})<\frac{1}{n}$.

This defines for each $m$ a subsequence $\{f_{\displaystyle n_{m,n}}\}_{n\in\mathbb{N}}$ of $f_n$. I taked $f_{\displaystyle n_{k,k}}=f_{n_k}$ then for each $k\in\mathbb{N}$ we have $\mu(x\in X: |f(x)-f_{n_k}(x)|\geq\frac{1}{k})<\frac{1}{k}$ so $f_{n_k}$ converges almost everywhere to $f$. I was quite informal in the last step, but this idea I think that works. How being more precise writting this last step, or is my idea completely wrong?

• Why does $f_{n_k}$ converge a.e. to $f$? This is the whole point. (In fact, I can construct functions $f_{n_k}$ and $f$ satisfying your condition so that$f_{n_k}$ do NOT converge a.e. to $f$, without passing to a further subsequence.) So I still suggest Borel-Cantelli. – GCD May 9 '13 at 14:55
• @GCD is my proof invalid? I'll really appreciate you if you can construct this subsequence or show that my "diagonal proof" not works. – Gaston Burrull May 9 '13 at 18:54

This got a bit long for a comment, but it continues our discussion from the comments above.

Up to your last step you have just used the definition of convergence in measure, and that is fine. But in the last step, you've hidden the main difficulty of the problem in your assertion, without proof, that $f_{n_k}$ converges a.e. to $f$.

In fact, your condition on $f_{n_k}$ is not enough to guarantee this.

Look at the measure space $[0,1]$ with Lebesgue measure. Let $f_k$ be the characteristic function of the interval $I_k$, where $I_k$ is an interval of length $1/k$ which move left along $[0,1]$ and "wrap around" when they reach the end, i.e. $$I_1 = [0,1]$$ $$I_2 = [0,1/2]$$ $$I_3 = [1/2,5/6]$$ $$I_4 = [5/6,1] \cup [0,1/12]$$ $$I_5 = [1/12,17/60]$$ etc.

(I hope you get the idea; try drawing a picture.)

Then $$\mu(\{x: |f_k(x) - 0| \geq 1/k\}) \leq 1/k$$ because each $f_k$ is $0$ except on an set of measure $1/k$.

On the other hand $\{f_k\}$ does not converge pointwise to zero anywhere, much less almost everywhere. (This is because the harmonic series diverges, and so any $x\in[0,1]$ is in some $I_k$ for infinitely many values of $k$.)

So my functions $f_k$ satisfy the exact same assumption as your $f_{n_k}$ do, but they do not converge pointwise almost everywhere.

So you need to refine your argument; the idea is to choose your $f_{n_k}$ better and then apply Borel-Cantelli.

• Nice example, I noted also your $f_{n_k}$ also converges in measure to $0$. – Gaston Burrull May 9 '13 at 22:47
• Yes, they certainly do. And a subsequence converges pointwise a.e. to zero, as it should, even though the full sequence does not. – GCD May 10 '13 at 1:34
• Where I find the so called "borel-cantelly answer"? – Gaston Burrull May 10 '13 at 1:35
• terrytao.files.wordpress.com/2012/12/… contains the theorem. I suggest you try to prove the Borel-Cantelli theorem without using any of the convergence theorems, as asked in an exercise. To use it you will have to choose some subsequence such that the "region of disagreement" shrinks fast enough. Intuitively, it prevents those disagreement regions from repeatedly covering any point except for a set of measure zero. – user21820 Nov 29 '13 at 3:28