Consider a measure space $(X, \mathcal{M}, \mu)$ with $\mu(X)<\infty$. Let $f_n$ be sequence of measurable functions such that $f_n \to f$ in measure. That is, for any $\varepsilon >0$, there exists $N$ such that for $n \geq N$, $$ \mu(\{x \in X: |f_n(x) -f(x)| \geq \varepsilon\}) < \varepsilon. $$ In addition, since $\mu(X)<\infty$, we also have for any measurable functions $g$, there exists a constant $M<\infty$ such that $$ \mu(\{x \in X: |g(x)| \geq M\}) < \varepsilon. $$ and below is my main question:
I was wondering for $n \geq N$, do I have the following set inclusion: $$ \left\{x: |g(x) (f_n(x) - f(x))| \geq \varepsilon \right\} \subset_? \left\{x: |g (x)| \geq M\right\} \cup \left\{x: |f_n(x) - f(x)| \geq \frac{\varepsilon}{ M} \right\} $$ I intended to believe the inclusion above holds but not very clear on it and would appreciate if someone could guide me on how to prove this or point out that I am doing something wrong. Thanks.
if $|g(x)| < M$ and $|f_n(x) - f(x)| < \frac\epsilon M$ then $|g(x) (f_n(x) - f(x))| < \epsilon$. Now take the complementary set