Convergence in measure of $\chi_{[n,\infty]}$

Question

If $f_n:=\chi_{[n,\infty)}$ with $f_n:\mathbb{R}\to\mathbb{R}$, then $\{f_n\}$ do not converge in measure.

Clearly $\{f_n\}$ do not converge to $f=0$ in measure. How can I prove that $\{f_n\}$ do not converge to any $f:\mathbb{R}\to\mathbb{R}$ in measure?.

Answer 1

Use that if $f_n \to g$ in measure and $f_n \to h$ pointwise, then $g = h$ a.e. This follows easily since convergence of $f_n$ to $g$ in measure implies pointwise a.e. to $g$ convergence along a subsequence $f_{n_k}$. But any subsequence of $f_n$ must also converge pointwise a.e. to $h$ and thus $$g(x) = \lim_{k\to\infty} f_{n_k}(x) = h(x) \,\,\, \text{ for a.e. } x \in \mathbb R.$$ That is, if both limits exist, they must be equal. Now show that $f_n \to 0$ pointwise and $f_n \not \to 0$ in measure and thus $f_n$ must not converge in measure.

Same rationale goes for almost uniform convergence since pointwise a.e. convergence implies almost uniform convergence on compact sets (Egorov's theorem).

Answer 2

Suppose $\{f_n\} \to f$ in measure. Fix $T \in (0,\infty)$. Note that $0 \leq x \leq T$ and $|f(x)|>\epsilon$ implies $|f_n(x)-f(x)| <\epsilon$ provided $n >T$. Hence $\mu \{0 \leq x \leq T$ and $|f(x)|>\epsilon\}=0$ for every $T$ and $\epsilon$. Clearly this implies $f=0$ almost everywhere.