# Composition of measurable maps is measurable

From Maths

TODO: write this - as it's mostly copied from the measurable map page

## Statement

Given two measurable maps their composition is measurable^{[1]}:

- [ilmath]f:(A,\mathcal{A})\rightarrow(B,\mathcal{B})[/ilmath] is measurable (same as saying: [ilmath]f:A\rightarrow B[/ilmath] is [ilmath]\mathcal{A}/\mathcal{B} [/ilmath]-measurable) and
- [ilmath]g:(B,\mathcal{B})\rightarrow(C,\mathcal{C})[/ilmath] is measurable

then:

- [ilmath]g\circ f:(A,\mathcal{A})\rightarrow(C,\mathcal{C})[/ilmath] is measurable.

In effect:

- [ilmath]\mathcal{A}/\mathcal{B} [/ilmath]-measurable followed by [ilmath]\mathcal{B}/\mathcal{C} [/ilmath] measurable [ilmath]=[/ilmath] [ilmath]\mathcal{A}/\mathcal{C} [/ilmath]-measurable

## Proof

TODO: See^{[1]} page 6 if help is needed (it wont be)

## References

- ↑
^{1.0}^{1.1}Probability and Stochastics - Erhan Cinlar