$\mu(\bigcup_{n=N}^\infty\omega:|g_n(\omega)-g(\omega)|>\varepsilon)<\varepsilon$ then $g_n\to g$ pointwise a.e.

Question

Let $(\Omega, \mathcal{A}, \mu)$ be a measure space and $g_n : \Omega \to \bar{\mathbb{R}}, n \in \mathbb{N}$ and $g : \Omega \to \bar{\mathbb{R}}$ be measurable functions.

Prove that if $$\forall \varepsilon,\exists N:\mu(\bigcup_{n=N}^\infty\{\omega:|g_n(\omega)-g(\omega)|>\varepsilon\})<\varepsilon$$

then $g_n \to g$ pointwise almost everywhere.

Could someone help me out?

Answer 1

To begin with notice that \begin{align*} &\bigcup_{n=N}^\infty \{ \omega : |g_n(\omega) - g(\omega)| > \varepsilon \} = \{ \omega : \sup_{n\geq N}|g_n(\omega) - g(\omega)| > \varepsilon \}\\ &\supset \{ \omega : \limsup_{n\to +\infty}|g_n(\omega) - g(\omega)| > \varepsilon \} \supset \{ \omega : \limsup_{n\to +\infty}|g_n(\omega) - g(\omega)| > 0 \}\\ &=\{ \omega : g_n(\omega) \not\to g(\omega) \}. \end{align*} Thus by the monotonicity of $ \mu $ $$ \mu ( \{ \omega : g_n(\omega) \not\to g(\omega) \} ) < \varepsilon, $$ which completes the proof.