# equality inverse image of a the supremum of sequence of function

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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Shouldn't there be $\bigcap$ instead of $\bigcup$? –  Martin Sleziak Apr 5 '11 at 16:30

I assume that the right hand side is an intersection.

To show the two sets are equal, show that every element in the set on the right hand side appears in the set on the left hand side, and vice versa.

Suppose that $x\in \{\omega:g(\omega)\leq r\}.$ This is equivalent to the condition $g(x)\leq r$. Since $g=\sup_{n} f_n$ we must have that $f_n(x)\leq r$ for every $n$ so that $x\in \cap_{n=1}^\infty \{\omega:f_n (\omega)\leq r\}.$

For the reverse inclusion, assume that $x\in\cap_{n=1}^\infty \{\omega:f_n (\omega)\leq r\}.$ Then for every $n$ we have that $f_n (x)\leq r$. This implies that $\sup_{n} f_n (x)\leq r$ and hence $g(x)\leq r$ so that $x\in \{\omega:g(\omega)\leq r\}$.

Hope that helps,

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The formula you give is wrong. It should read $$\{\omega\,:\,g(\omega)\leq r\} = \bigcap_{n=1}^{\infty} \{\omega\,:\,f_{n}(\omega) \leq r\}.$$ To prove this, note: If $A \subset B$ and $A \supset B$ then $A = B$. Equivalently: if for all $a \in A$ we have $a \in B$ and if for all $b \in B$ we have $b \in A$ then it follows that $A = B$.

Note that $\omega \in \bigcap_{n=1}^{\infty} \{\omega\,:\,f_{n}(\omega) \leq r\}$ means that for all $n$ we have $f_{n}(\omega) \leq r$, so $g(\omega) = \sup_{n} f_{n}(\omega) \leq r$ as well. This shows $\supset$ above. Conversely, if $g(\omega) \leq r$ then for all $n$ we have $f_{n}(\omega) \leq g(\omega) \leq r$ by the definition of the supremum, so $\omega \in \bigcap_{n=1}^{\infty} \{\omega \,:\, f_{n}(\omega) \leq r\}$ and this shows $\subset$ above.

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