# Why does $L^2$ convergence not imply almost sure convergence

What's wrong with this argument?

Let $f_n$ be a sequence of functions such that $f_n \to f$ in $L^2(\Omega)$. This means $$\lVert f_n - f \rVert_{L^2(\Omega)} \to 0,$$ i.e., $$\int_\Omega(f_n - f)^2 \to 0.$$ Since the integrand is positive, this must mean that $f_n \to f$ a.e.

Why is this not true? Apparently this only true for a subsequence $f_n$ (and in all $L^p$ spaces).

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Can you provide a proof of the fact: $$\int_{\Omega}(f_n-f)^2\to 0\implies f_n\to f?$$ – leo Jul 5 '12 at 21:07
$\int_\Omega (f_n - f)^2$ being small doesn't mean $(f_n - f)^2$ is small everywhere. It could be large on a set of small measure. And for different $n$ those sets of small measure could hit any particular point infinitely often. – Robert Israel Jul 5 '12 at 21:07
Look at this for a counterexample – leo Jul 5 '12 at 21:09
@leo Well, don't we say that $\lVert f_n - f \rVert_{L^2(\Omega)} = 0$ implies $f_n -f = 0$ a.e. This is similar..? – blahb Jul 5 '12 at 21:09
Yes, but you do not have $||f_n-f||=0$, you have $$||f_n-f||\to 0.$$ – leo Jul 5 '12 at 21:11

Consider the following sequence of characteristic functions $f_n \colon [0,1] \to R$ defined as follows:

$f_1 = \chi[0, 1/2]$

$f_2 = \chi[1/2, 1]$

$f_3 = \chi[0, 1/3]$

$f_4 = \chi[1/3, 2/3]$

$f_5 = \chi[2/3, 1]$

$f_6 = \chi[0, 1/4]$

$f_7 = \chi[1/4, 2/4]$

and so on.

Then $f_n \to 0$ in $L^2$, but $f_n$ does not converge pointwise.

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