# [SOLVED]Measurable Function

##### Active member
Another analysis review question:

Suppose that $$\displaystyle f:\mathbb{R}\rightarrow\mathbb{R}$$ is a measurable function and that $$\displaystyle g:\mathbb{R}\rightarrow\mathbb{R}$$ is a Borel (i.e. Borel measurable) function. Show that $$\displaystyle f\circ g$$ is measurable.

If we only assume that g is measurable, is it still true that the composition $$\displaystyle f\circ g$$ is measurable?

##### Active member
I'm fairly confident in my proof that $$\displaystyle f\circ g$$ is measurable in the instance that g is Borel.

Note that a suitable definition of measurability is that f is measurable iff $$\displaystyle f^{-1}(U)$$ is measurable for an arbitrary open set U in $$\displaystyle \mathbb{R}$$. A similar definition can be written for a Borel function, i.e. that g is Borel iff $$\displaystyle g^{-1}(U)$$ is a Borel set for an arbitrary open set U in $$\displaystyle \mathbb{R}$$.

That being said, the first proof amounts to using the fact that the measurable sets form a sigma-algebra, and we may write the pull-back of a Borel set as the arbitrary union, intersection, and complement of the pull-back of open sets.

The second part is tricky. What I need to know is whether for an arbitrary measurable set $$\displaystyle E\subset\mathbb{R}$$ and a measurable function f, we have $$\displaystyle f^{-1}(E)$$ is measurable. My intuition is that this should not be the case, but finding a suitable counter-example has proven to be difficult.

#### Evgeny.Makarov

##### Well-known member
MHB Math Scholar
See the Wikipedia page about the composition of measurable functions, and see StackExchange for a counterexample concerning the composition of two Lebesgue-measurable functions.