I want to show that if $f_n $ converges in measure to $f $ and for each $n$ $\mu ( \{x : |\hat {f _n (x)} -f_n(x) \ge \frac {1 } {n } \} )=\frac {1 } {n } $, then $\{\hat {f _n }\} $ converges in measure to $f$.
A sequence $\{f _n \} $ of measurable functions is said to converge in measure if there is a measurable function $f $ such that, for any $\epsilon >0$
$$\lim \mu ( \{x : |f _n (x)-f(x) \ge \epsilon \} )=0$$
I sort of know that since $|\hat {f _n } -f |\le |\hat {f _n } -f _n |+ |f _n - f |$, then
$\{x: |\hat {f _n }(x) -f(x) |\ge \epsilon \} \subset \{x:|\hat {f _n }(x) -f _n(x) |+ |f _n(x) - f(x) |\ge \epsilon \} $.
But how can I claim that $\mu (\{x: |\hat {f _n }(x) -f(x) |\ge \epsilon \}) \le \mu ( \{x:|\hat {f _n }(x) -f _n(x) |\ge \epsilon\} )+ \mu(\{x:|f _n(x) - f(x) |\ge \epsilon \}) $
Thanks in advance!
Since $|\hat f_n(x) - f(x)| \le |f_n(x) - f(x)| + |\hat f_n(x) - f_n(x)|$, at least one of the terms on the right must be greater than $\frac{\epsilon}{2}$ if the expression on the left is greater than $\epsilon$. That is, $$ \{ x : |\hat f_n(x) - f(x)| > \epsilon \} \subset \{ x : |f_n(x) - f(x)| > \epsilon/2 \} \cup \{ x : |\hat f_n(x) - f_n(x)| > \epsilon/2 \}.$$ Then $$ \mu(\{| \hat f_n - f| > \epsilon \}) \le \mu(\{ |f_n - f| > \epsilon/2\}) + \mu(\{ |\hat f_n - f| > \epsilon/2\}).$$ Both terms on the right tend to 0 as $n \to \infty$.