Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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).

share|cite|improve this question
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
up vote 11 down vote accepted

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.