Convergence in measure

Remarks

Example: on $[0, 1]$ with Lebesgue measure, write $n = 2^k + j$ with $0 \le j < 2^k$ and let $f_n = \mathbf{1}_{[j 2^{-k}, (j+1) 2^{-k}]}$. For $0 < \varepsilon \le 1$, the set where $\abs{f_n} \ge \varepsilon$ has measure $2^{-k}$, which tends to $0$, so $f_n \to 0$ in measure. But each $x \in [0, 1]$ has $f_n(x) = 1$ for infinitely many $n$, so $f_n(x)$ converges for no $x$. Non-example: on $\R$ with Lebesgue measure, $f_n = \mathbf{1}_{[n, n+1]}$ tends to $0$ at every point, but the set where $\abs{f_n} \ge 1/2$ has measure $1$ for all $n$. A local variant asks the limit to be $0$ only for the restriction to each set of finite measure. Convergence in $L^p$ for some $1 \le p < \infty$ implies convergence in measure, by Chebyshev's inequality.