If $f_k\to f$ in measure, does $f_k\to f$?

Question

I recall that

1) $f_k(x)\to f(x)$ if $\lim_{k\to \infty }f_k(x)=f(x)$

2) $f_k\to f$ in measure if $$\forall \varepsilon>0: \lim_{n\to \infty }m(\{x\mid |f_k(x)-f(x)|\geq \varepsilon\})=0$$

3) $f_k\to f$ in $L^1$ if $\lim_{k\to\infty }\int |f_k-f|=0$.

I proved that $1)\implies 2)$, that $3)\implies 2)$ and that $2)$ doesn't imply $3)$.

What about $2)\implies 1)$ and $3)\implies 1)$ ? I think that both are wrong, but I can't find counter example.

Answer 1

Convergence in measure does not imply pointwise convergence anywhere. A nice sequence that you can use to prove this is a sequence of indicator functions like the indicators of the following sequence of intervals:

\begin{align*} &[0, 1/2], [1/2, 1] \\ &[0, 1/3], [1/3, 2/3], [2/3, 1] \\ &[0, 1/4], ... \end{align*}

This converges to zero in measure and $L^1$, but does not converge pointwise anywhere on $[0, 1]$.

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The best you can get from $L^1$ convergence is pointwise convergence of a subsequence. This still works in this case.