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
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    Yes, just try out writing it through the definition of the measurability via the inverse of sets – Ilya Jan 21 '13 at 13:57
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    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
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    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 $k^{-1}(1) = (1_B \circ h)^{-1}(1) = h^{-1}(B) = g(B) = A$.

  • Should the last line read $k^{-1}(B) = (1_B \circ h)^{-1}(B) = h^{-1}(B) = g(B) = A$ – man_in_green_shirt Aug 17 '16 at 15:40
  • @man_in_green_shirt no because $1_B^{-1}(1)=B$ – John Cataldo Mar 21 at 17:30

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 $g \circ f$

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    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

As an addition to the other answers, I think it is worth noting that in terms of imitating the relation between a topology and a continuous function when thinking of the relation between a $\sigma$-algebra and a measurable function, we can only go so far as having $g$ Borel measurable.

Let $\mathcal{T}$ be the open sets, $\mathcal{B}$ be the Borel sets and $\mathcal{M}$ be the Lebesgue measurable sets of the real line, respectively. Then we have that $\mathcal{T}\subset\mathcal{B}\subset\mathcal{M}$.

We call $f:E\to\mathbb{R}$ continuous if $f^{-1}(\mathcal{T})\subseteq\mathcal{T}$, Borel measurable if $f^{-1}(\mathcal{T})\subseteq\mathcal{B}$ and Lebesgue measurable if $f^{-1}(\mathcal{T})\subseteq\mathcal{M}$, respectively. Accompanying the inclusions of the classes of sets, we have that every continuous function is Borel measurable and every Borel measurable function is Lebesgue measurable.

It turns out that $f$ is Borel iff $f^{-1}(\mathcal{B})\subseteq\mathcal{B}$, which is analogous to the continuous case, but we don't have $f^{-1}(\mathcal{M})\subseteq\mathcal{M}$ in general for Lebesgue measurable functions $f$. The best we can do is that $f$ is Lebesgue measurable iff $f^{-1}(\mathcal{B})\subseteq\mathcal{M}$.

Consequently if $f$ is Lebesgue measurable and $g$ is Borel measurable, then $g\circ f$ is Lebesgue measurable (just like in the continuous case), and if $f,g$ are both Borel measurable then $g\circ f$ is Borel measurable (again, just like in the continuous case), but if both $f,g$ are Lebesgue measurable we can cook up examples where $g\circ g$ is not Lebesgue measurable. (For the sake of completeness, if $g$ is Lebesgue and $f$ is Borel then $g\circ f$ need not be Lebesgue: in the standard example above f is a homeomorphism).

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