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?
[Math] Convergence in measure does not imply $L^1$ convergence
measure-theory
Best Answer
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.