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Suppose that $f_n$ is a sequence of real measurable functions on a set $X$ of finite measure, and suppose that there is some $\epsilon$ such that for all $n\geq1$:


I want to show that this is false: $f_n(x)\rightarrow f(x)$ a.e. on $X$.

I tried to prove the existence of a set $Y\subseteq X$ of positive measure such that $f_n(x)$ is far from $f(x)$ for all $x\in Y$. However, I only obtain sets $Y_n$ that depend on $n$ so I guess I need some stronger argument. Any help is appreciated!

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If $f_n \to f$ a.e. then for all $\epsilon > 0$ you have $$m \{ x\ :\ \lvert f_n(x)-f(x)\rvert \ge \epsilon \} \to 0.$$ Have a look at this: – Giuseppe Negro Mar 22 '12 at 17:37
This would work but I don't see why it is true :( – wircho Mar 22 '12 at 17:55
Note that your initial idea is bound to fail, consider for instance $f(x) = 1$, $f_n(x) = (-1)^n(\chi_{[-1,0]}(x) - \chi_{(0,1]}(x))$ on $X = [-1,1]$. What you really want to prove is that the set $$Y = \{x\in X\mid \limsup_n |f_n(x) - f(x)| \ge \epsilon\}$$ has positive measure. – Sam Mar 22 '12 at 17:55
Hint: Show that a point $x\in X$ is in $Y$ if and only if it is in $\bigcup_{n=m}^\infty Y_n$ for each $m\in \mathbb Z_+$. Here $Y_n = \{x: \; |f_n(x) - f(x)|\ge \epsilon\}$ and use this to show $m(Y) > 0$. – Sam Mar 22 '12 at 17:58
up vote 2 down vote accepted

We define the convergence in measure:

A sequence $\{f_n\}$ converges in measure to $f$ if for all $\delta>0$ we have $$\lim_{n\to\infty}m(\{|f_n(x)-f(x)|\geq\delta\})=0.$$

We have the following result:

Let $(X,\mathcal B,\mu)$ a finite measured space. If the sequence $\{f_n\}$ of measurable functions converges almost everywhere to $f$, then it converges in measure to $f$.

To see this,fix $\delta>0$ and note that $m\left(\bigcap_k\bigcup_{n\geq k}\{x:|f_n(x)-f(x)|\}\right)=0$. Since the sequence $\left\{\bigcup_{n\geq k}\{x:|f_n(x)-f(x)|\}\right\}_k$ is decreasing and we are in a finite measured space we have $m\left(\bigcup_{n\geq k}\{x:|f_n(x)-f(x)|\}\right)$ converges to $0$ and we get the result.

Note that the fact that the space is finite is primordial, since you get a counter-example taking the real line, with the Borel $\sigma$-algebra and Lebesgue measure, taking $f_n$ equal to the characteristic function of $(n,n+1)$.

Now, to solve the problem note that the sequence $\{f_n\}$ fail to converge in measure.

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