If F Measurable and G Continuous Than F Composition G is Measurable

Is composition of measurable functions measurable?

Solution 1

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 $k^{-1}(1) = (1_B \circ h)^{-1}(1) = h^{-1}(B) = g(B) = A$.

Solution 2

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.

Solution 3

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 measurable spaces and $f:E_1 \to E_2$ and $g:E_2 \to E_3$ measurables functions.If $X \in \mathcal T_3$, then , by measurability of $g$ we have : $g^{-1}(X) \in \mathcal T_2$ and by measurability 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 measurability of $g \circ f$

Solution 4

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).

Comments

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.
- Yes, just try out writing it through the definition of the measurability via the inverse of sets
- 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.
- 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.
As A.D. notes, this is probably not what the OP means. Since he now mentions "Lebesgue measurable function".
Should the last line read $k^{-1}(B) = (1_B \circ h)^{-1}(B) = h^{-1}(B) = g(B) = A$
@man_in_green_shirt no because $1_B^{-1}(1)=B$
Hi! Thank you for the explanation, it is a great complement in order to get the thing. Please, can you tell me why the case M to M is not working for lebesgue measurable? Is this a too light restriction for such functions? Are there too many such functions?
To the contrary, it's too strict, even at the syntactic level: indeed, since $\mathcal{B}\subsetneq\mathcal{M}$, $\{f|f^{-1}\mathcal{M}\subseteq\mathcal{M}\}\subsetneq\{f|f^{-1}\mathcal{B}\subseteq\mathcal{M}\}$.
Here is one way to think about this: By the standard example above presented by Mirjam we know that there are homeomorphisms of closed intervals that fail to take a leb-measurable set to another leb-measurable set under preimages. Thus if we defined Lebesgue measurability of $f$ as the requirement that $f^{-1}\mathcal{M}\subseteq \mathcal{M}$, we would have to be content with continuous functions not being measurable; which in the standard way of developing integrals would prohibit one to integrate a continuous function; which in turn means that we would have to let calculus go.
(There is nothing wrong with letting calculus go per se, though unless you have a better idea the number of your followers will be very limited.)

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Source: https://9to5science.com/is-composition-of-measurable-functions-measurable

Robson Culdriatat