Convergence in measure does not imply $L^1$ convergence 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?
 A: 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.
A: 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.
