3
$\begingroup$

can somebody explain me this equality please

{ω : g(ω) ≤ r} =$\bigcup^\infty_{n=1}${ω : $f_{n}$(ω) ≤ r}

where $g=\sup_{n\geq 1} f_n$

this is to prove that the supremum of mesurable sequences is mesurable too but i can't understand where the equality came from.

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