# I'm looking for a counterexample - where a sequence converges pointwise and in measure, but not almost uniformly.

I'm looking for a counterexample - where a sequence converges pointwise and in measure, but not almost uniformly.

There is a theorem (which I don't know how to prove) which states that if $|f_n| \leq g \in L^1$ and $f_n \rightarrow f$ pointwise and in L^1, then $f_n \rightarrow f$ almost uniformly.

$f_n \rightarrow f$ in $L^1$ implies in measure. I want to abandon $L^1$ convergence and only focus on convergence in measure. Can anyone advise me to the end of this proof, or show me an example of a sequence which converges almost everywhere, in measure, but not almost uniformly?

Thanks.

• Your "theorem" seems strange. Pointwise convergence of $(f_n)$ and the boundedness condition imply $L_1$ convergence. Commented Jan 21, 2014 at 0:05
• Here's a link to a post that contains your theorem and its proof, I believe. Commented Jan 21, 2014 at 1:22

I think, working over $\Bbb R$, the sequence $(f_n)$ defined by
$\ \ \ \ \ \ \$$f_1=\chi_{[0,1]}$, $f_2=\chi_{[1,3/2]}$, $f_3=\chi_{[3/2,2]}$, $f_4=\chi_{[2, 2+1/4]}$, $f_5=\chi_{[2+1/4,2+1/2]}$, $\ldots$
will work (I hope it's obvious how to define the rest of the $f_n$).
$(f_n)$ converges pointwise and in measure to $0$. But, given any set $E$ of measure less than $1/2$ and any $N$, there is an $x\notin E$ and an $n>N$ with $f_n(x)>1/2$; so the sequence in not almost uniformly convergent.
(On the other hand, I'm fairly certain that a sequence $(f_n)$ that converges in measure has a subsequence that converges almost uniformly. The proof eludes me at the moment, but I'll think about it...)