Show that $\left\{ \omega \mid sup_{n}\{h_n(\omega)\} > c \right\} = \bigcup_{n=1}^{\infty} \left\{ \omega \mid h_n (\omega) > c \right\}$ Given a sequence of measurable functions $h_n (\omega)$ I manged to prove that the sequence $inf_{n}\\{h_n(\omega)\\}$ is also measurable. For obtaining the same result for $sup_{n}\\{h_n(x)\\}$ I need to show that $$\left\\{ \omega \mid sup_{n}\\{h_n(\omega)\\} > c \right\\} = \bigcup_{n=1}^{\infty} \left\\{ \omega \mid h_n (\omega) > c \right\\}$$ I managed to show that the left-hand set in contained in the right-hand set, however I cannot show the converse. Can someone give a detailed demonstration of the latter?

If $\omega$ is in the union, then $h_{n_0}(\omega)>c$ for some $n_0$, hence $\sup_nh_n(\omega)\geq h_{n_0}(\omega)>c$. This shows that the right-hand side is contained in the left-hand side.