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I'm supposed to find a counterexample to the statement to:

"Lemma": If $f_n \to f$ in $L^1([0,1])$, then $f_n(x) \to f(x)$ for almost every $x \in[0,1]$.

What's wrong with the proof below for this "Lemma"?

"False Proof": We know that $||f_n-f||_1\to0$ as $n\to\infty$. Let $\epsilon > 0$ and $n\in\mathbb{N}$, and consider $B_{n,\epsilon} = \{x\in[0,1]:|f_n(x)-f(x)|\geq \epsilon\}$. Then integrating $|f_n - f|$ over $B_{n,\epsilon}$, we get $m(B_{n,\epsilon})\leq\frac{1}{\epsilon}||f_n-f||_1$. So fixing $\epsilon$, we get $\lim_{n\to\infty}m(B_{n,\epsilon})=0.$ Thus, the set $X_\epsilon$ that consists of $x\in[0,1]$ that are in $B_{n,\epsilon}$ for infinitely many $n$ has measure $0$. So $X=\cup_{i=1}^\infty X_{1/i}$ has measure $0$. For $x$ outside of $X$, for each $\epsilon = 1/i$, we have that $x \notin X_\epsilon$, so there exists $N$ s.t. for all $n > N$, $|f_n(x) - f(x)| < \epsilon$, which implies $\lim_{n\to\infty}f_n(x) = f(x)$

To put the question succinctly, what's the problem in here, and what would a counterexample be? Thanks!

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    @countinghaus: You could turn that into an answer.2012-04-30

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The answer was given in comments: I add an illustration under them.

Thus, the set $X_\epsilon$ that consists of $x\in [0,1]$ that are in $B_{n,\epsilon}$ for infinitely many $n$ has measure $0$." This is your error. Consider covers of $[0,1]$ by a single interval, by two intervals of length $1/2$, by three intervals of lengths $1/3$ etc. Measures of these go to $0$, but each $x$ lies in infinitely many members of the covers.

– Miha Habič

Taking characteristic functions of Miha's sets, properly enumerated, gives you the counterexample!

– countinghaus

On the graph below, I multiplied characteristic functions by different constants close to $1$, to make the graph more legible. The functions can be put into a sequence in the order of their height on the picture: shortest first.

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