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[SOLVED] Measurable Function

TheBigBadBen

Active member
May 12, 2013
84
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?
 

TheBigBadBen

Active member
May 12, 2013
84
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
Jan 30, 2012
2,493
See the Wikipedia page about the composition of measurable functions, and see StackExchange for a counterexample concerning the composition of two Lebesgue-measurable functions.
 

TheBigBadBen

Active member
May 12, 2013
84
The Stack Exchange counterexample was exactly what I was looking for. You have helped me tremendously. Thank you.