We know that for a sequence of measurable functions, $L^p$ convergence implies convergence in measure. However, the converse is false. I'm having trouble coming up with a counter-example, please help! Also, does convergence in measure differ from a.e. pointwise convergence?

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    $\begingroup$ Convergence in measure does not imply pointwise convergence almost everywhere. One counterexample is a sequence of characteristic functions of intervals of length converging to 0 such that every point is both in and out of infinitely many intervals. For instance, the intervals could be $[0,1/2],[1/2,1],[0,1/4],[1/4,1/2],...$. $\endgroup$ Oct 3, 2012 at 22:15

2 Answers 2


Convergence almost everywhere implies convergence in measure for finite measure space. You can see this by applying the dominated convergence theorem to the integral of the function $\chi_{\{|f_n - f|>\epsilon\}}$ (which is the measure of the set $\{|f_n - f|>\epsilon\}$), which is dominated by $1$ on a finite measure space. On an infinite measure space, this need not hold. An example where it fails is $f_n (x) = \chi_{[n,\infty)}(x)$ which converges to zero pointwise everywhere, but for $\epsilon < 1$ we have $\mu\{x: |f_n(x)|>\epsilon\} = \infty$ for any $n$, so it cannot converge in measure.

An example of a function which converges to zero in measure but not in $L^1$ is $f_n = n \chi_{[0,\frac{1}{n}]}$, since $\int |f_n| = 1$ for each $n$, but $\mu\{|f_n| > \epsilon\} = \frac{1}{n}$ for $\epsilon$ small.


Here is a counterexample such that we have convergence in measure, but not in $L^p$. Take $f_n=2^{-n}1_{[2^n,2^{n+1}]}$, then $|\{f_n>\epsilon\}|\rightarrow 0$. Hence $f_n\rightarrow 0$ in measure, but $||f_n||_{L^p}=1$ for all $n$.

EDIT: My previous counterexample was not correct, and I made a false statement about convergence a.e. I have corrected my counterexample, and removed the false statement.

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    $\begingroup$ Surely $m \{x | |f_n(x) - 0| \geq \epsilon \} = 1$ for all $\epsilon > 0$? Maybe $f_n = n 1_{[n,n+\frac{1}{n}]}$? Then $f_n(x) \to 0 $ a.e., $\|f_n\|_1 = 1$, and $f_n$ converges to $0$ in measure. $\endgroup$
    – copper.hat
    Oct 3, 2012 at 22:22
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    $\begingroup$ Thank you very much. I was sloppy, and too quick to answer. I mostly work in probability spaces, hence finite measure spaces, so I did not proof my work correctly. My new answer should be correct. Sorry for the confusion. $\endgroup$ Oct 3, 2012 at 23:29

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