Measurable Function Composition: f∘g

[TheBigBadBen]

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]

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]

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]

The Stack Exchange counterexample was exactly what I was looking for. You have helped me tremendously. Thank you.

 

[blue_raver22]

The composition of two measurable functions, f and g, is also measurable. This is because the composition of two measurable functions results in a new function that maps measurable sets to measurable sets. This can be proven by considering the preimage of a measurable set under the composition function f∘g. Since g is Borel measurable, the preimage of any Borel measurable set will also be Borel measurable. Then, since f is measurable, the preimage of the Borel measurable set under f will also be measurable. Therefore, the composition function f∘g maps measurable sets to measurable sets, making it measurable.

If we only assume that g is measurable, it is not necessarily true that the composition f∘g is also measurable. This is because the preimage of a measurable set under the composition function may not be measurable. For example, let g(x) = 1 for all x∈ℝ. This function is measurable since it maps all Borel measurable sets to the entire real line, which is measurable. However, let f(x) = 1 if x is rational and 0 if x is irrational. This function is not measurable since the preimage of any Borel measurable set under f will either be the entire real line (if the set contains a rational number) or the empty set (if the set contains only irrational numbers), neither of which are measurable. Therefore, the composition f∘g is not measurable.