# Is composition of measurable functions measurable?

We know that if $f: E \to \mathbb{R}$ is a Lebesgue-measurable function and $g: \mathbb{R} \to \mathbb{R}$ is a continuous function, then $g \circ f$ is Lebesgue-measurable. Can one replace the continuous function $g$ by a Lebesgue-measurable function without affecting the validity of the previous result?

-
Please, try to make the title of your question more informative. E.g., Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. –  Martin Sleziak Jan 21 '13 at 13:56
Yes, just try out writing it through the definition of the measurability via the inverse of sets –  Ilya Jan 21 '13 at 13:57
Do you mean Borel or Lebesgue measurable? More generally, when we deal with measurability, we always have to give precisions about the involved $\sigma$-algebras. –  Davide Giraudo Jan 21 '13 at 13:59
An equivalent formulation: The inverse image of a Lebesgue measurable set under a Lebesgue measurable function is Lebesgue measurable. Which is not the case in general. –  AD. Jan 21 '13 at 14:56

Here is the standard example:

Let $f\colon [0,1]\to [0,1]$ be the Cantor–Lebesgue function. This is a monotonic and continuous function, and the image $f(C)$ of the Cantor set $C$ is all of $[0,1]$. Define $g(x) = x + f(x)$. Then $g\colon [0,1] \to [0,2]$ is a strictly monotonic and continuous map, so its inverse $h = g^{-1}$ is continuous, too.

Observe that $g(C)$ measure one in $[0,2]$: this is because $f$ is constant on every interval in the complement of $C$, so $g$ maps such an interval to an interval of the same length. It follows that there is a non-Lebesgue measurable subset $A$ of $g(C)$ (Vitali's theorem: a subset of $\mathbb{R}$ is a Lebesgue null set if and only if all its subsets are Lebesgue measurable).

Put $B = g^{-1}(A) \subset C$. Then $B$ is a Lebesgue measurable set as a subset of the Lebesgue null set $C$, so the characteristic function $1_B$ of $B$ is Lebesgue measurable.

The function $k = 1_B \circ h$ is the composition of the Lebesgue measurable function $1_B$ and and the continuous function $h$, but $k$ is not Lebesgue measurable, since $k^{-1}(1) = (1_B \circ h)^{-1}(1) = h^{-1}(B) = g(B) = A$.

-

This Wikipedia article may be what you are looking for.

According to the article, a function $f: \mathbb{R} \to \mathbb{R}$ is said to be Lebesgue-measurable if and only if for every Borel-measurable subset $B$ of $\mathbb{R}$, its pre-image ${f^{\leftarrow}}[B]$ is a Lebesgue-measurable subset of $\mathbb{R}$. Therefore, because we are dealing with two different $\sigma$-algebras of $\mathbb{R}$ here, the composition of two Lebesgue-measurable functions is not necessarily Lebesgue-measurable. This is precisely what GEdgar and AD. have mentioned.

-

This details the Ilya comment:

Yes. In general let $(E_1,\mathcal T_l) ,(E_2,\mathcal T_2),(E_3,\mathcal T_3)$ be three meurable spaces and $f:E_1 \to E_2$ and $g:E_2 \to E_3$ mesurables functions.If $X \in \mathcal T_3$, then , by mesurability of $g$ we have : $g^{-1}(X) \in \mathcal T_2$ and by mesurability of $f$ we have $f^{-1}(g^{-1}(X)) \in \mathcal T_1$. Since $(g \circ f)^{-1} (X)=f^{-1} ( g^{-1}(X))$ we have :$$\forall X\in \mathcal T_3 \quad (g \circ f)^{-1} (X) \in \mathcal T_1$$
This gives mesurability of $g \circ f$

-
As A.D. notes, this is probably not what the OP means. Since he now mentions "Lebesgue measurable function". –  GEdgar Jan 21 '13 at 15:46