If $S=(X, \mathcal{F})$ is a measurable space, $\mu,\nu$ two measure on S s.t. $\nu\ll\mu$, are all $\mu$-measurable functions also $\nu$-measurable?

Question

Let $S = (X,\mathcal{F})$ be a measurable space, i.e. $X$ be a set and $\mathcal{F}$ be a $\sigma$-algebra on $X$. Suppose $\mu, \nu$ are two measures on $S$ such that $\nu$ is absolutely continuous with respect to $\mu$, i.e. write $\nu\ll\mu$ to denote that $\forall A\in \mathcal{F}:\mu(A) = 0\implies \nu(A) = 0$.

Is it then true that every $\mu$-measurable function $f$ is also $\nu$-measurable? That is, if $(Y, \mathcal{G})$ is another measurable space and $f:X\to Y$ is a map such that $\forall A \in G:\text{$f^{-1}(A)\in \mathcal{F}$ and $f^{-1}(A)$ is $\mu$-measurable}$.

This can very well be a completely trivial question, but I don't recall many tools that allow one to tackle a question like this. As of now, I also don't know how to approach it: Canonically we would like to show that

$$\forall A\in \mathcal{G}, B\subset X, : \nu(B) = \nu(f^{-1}(A)\cap B) + \nu(B\setminus f^{-1}(A))$$

and our only source of information is that

$$\forall A\in \mathcal{G}, B\subset X, : \mu(B) = \mu(f^{-1}(A)\cap B) + \mu(B\setminus f^{-1}(A))$$

How should we apply the absolute continuity in this case?

Answer 1

First, some remarks on measurability

There are several different notions of "$\mu$-measurable function" that can be found in the litterature. In my opinion, there are three "main" important ones and they happen to be closely related in the case of functions with values in a metric space.

Concretely, if for every subset $A$ of $X$ you let $\mu^\ast(A)=\inf\lbrace\mu(B)~|~B\in\mathcal F\text{ and }A\subset B\rbrace$ (i.e. $\mu^\ast$ is the outer measure associated to $\mu$) and if you you endow $X$ with the coarsest uniform structure making every $\mathbf R$-valued $\mathcal F$-simple function on $X$ uniformly continuous, then given a function $g$ defined on $X$ taking its values in a metric space $Z$, the following conditions are equivalent:

1. for every $K\in\mathcal F$ such that $\mu(K)<+\infty$, there exists a coutanble subset $Q$ of $Z$ such that for $\mu$-almost every $x\in K$, the point $g(x)$ of $Z$ is adherent in $Z$ to $Q$, and for every Borel subset $U$ of $Z$ and every subset $B$ of $X$, $$\mu^\star(B)=\mu^\star(B\cap g^{-1}(U))+\mu^\star(B\setminus g^{-1}(U)),$$
2. for every $K\in\mathcal F$ such that $\mu(K)<+\infty$, there exists a sequence $(g_n)_{n\in\mathbf N}$ of $Z$-valued $\mathcal F$-simple functions on $K$ such that for $\mu$-almost every $x\in K$, the sequence $(g_n(x))_{n\in\mathbf N}$ of elements of $Z$ converges in $Z$ to $g(x)$,
3. for every $K\in\mathcal F$ such that $\mu(K)<+\infty$ and every $\varepsilon>0$, there exists a $L\in\mathcal F$ contained in $K$ such that $\mu(K\setminus L)\leq\varepsilon$ and such that the restriction of $f$ to $L$ is uniformly continuous.

Proving $(1.\Leftrightarrow 2.)$ is not very difficult. The implication $(2.\Rightarrow 3.)$ is a consequence of Egorov's theorem. Proving $(3.\Rightarrow 1.)$ or $(3.\Rightarrow 2.)$ is, I think, a bit harder…

When $g$ satisfies these equivalent conditions, it is customary to say that $g$ is $\mu$-measurable. And if $A$ is a subset of $X$, it is customary to say that $A$ is $\mu$-measurable if its indicator function is $\mu$-measurable. This notions of $\mu$-measurability for sets is none other than the notion of measurability associated to the outer measure $\mu^\star$ in the Carathéodory's extension theorem.

If I understood your question correctly from your edit, you intended to use this notion of measurability for sets and your question basically boils down to "are $\mu$-measurable subsets of $X$ necessarily $\nu$-measurable?".

A negative answer

If $\mu$ is given by $\mu(\varnothing)=0$ and $\mu(A)=+\infty$ if $A\in\mathcal F\setminus\{\varnothing\}$, it is easy to see that every subset of $X$ is $\mu$-measurable and every measure on defined on $\mathcal F$ is absolutely continuous with respect to $\mu$. So if you know of an example of a measure $\nu$ for which there are non-$\nu$-measurable subsets of $X$, you can see that the answer to your question in its full generality is no.

For example, if $X$ is the real line $\mathbf R$, $\mathcal F$ the $\sigma$-algebra of Lebesgue-measurable sets, $\nu$ the Lebesgue measure and $\mu$ given as above, then $\nu\ll\mu$ and any Vitali set is $\mu$-measurable but not $\nu$-measurable.

A positive answer

There are many practical situations in which one can answer positively to your question.

If for every $K\in\mathcal F$ such that $\nu(K)<+\infty$ you can find a partition of $K$ made of a set $N\in\mathcal F$ such that $\nu(N)=0$ and of sets $K_0$, $K_1$, $K_2$, … $\in\mathcal F$ such that $\mu(K_n)<+\infty$ for every $n\in\mathbf N$, then it is not difficult using condition 2. to prove that $\mu$-measurable sets are $\nu$-measurable.

Indeed, suppose that $A$ is a $\mu$-measurable subset of $X$. Let $K\in\mathcal F$ such that $\nu(K)<+\infty$. Write $K=N\cup K_0\cup K_1\cup K_2\cup\cdots$ as above. Since $A$ is $\mu$-measurable, there exists for every $n\in\mathbf N$ a sequence $(g_{n,m})_{m\in\mathbf N}$ of $\mathbf R$-valued $\mathcal F$-simple functions on $K_n$ that converges $\mu$-almost everywhere in $K_n$ to the indicator function of $A$. If for every $x\in K$ and every $n,m\in\mathbf N$, you let $$g_m(x)=\begin{cases}0&\text{if }x\in N\text{ or if }x\in K_n\text{ and }m<n,\\ g_{n,m}(x)&\text{if }x\in K_n\text{ and }m\geq n,\end{cases}$$ then $(g_m)_{m\in\mathbf N}$ is a sequence of $\mathbf R$-valued $\mathcal F$-simple functions on $K$ that converges $\mu$-almost everywhere on $K$ to the indicator function of $A$, and since $\nu\ll\mu$, $\mu$-almost everywhere is also $\nu$-almost everywhere.

This kind of partition is always possible in the following very common situations:

• $\mu$ is $\sigma$-finite,
• $\nu$ has a density with respect to $\mu$,
• $X$ is a Hausdorff topological space, $\mathcal F$ contains its Borel $\sigma$-algebra and $\mu$ and $\nu$ are Radon measures.

Answer 2

Let $(X, \mathcal F)$ and $(Y, \mathcal G)$ two measurable spaces and $f: X \to Y$. We call $f$ measurable if $\forall E \in \mathcal G$, $f^{-1}(E) \in \mathcal F$.

You see that the definition does not involve any measure such that $\mu(f^{-1}(E))$ and $\nu(f^{-1}(E))$ are well defined for every $E \in \mathcal G$ as both measures are defined on the same $\sigma$-algebra.

Intuitively speaking, the set $(X, \mathcal F)$ is called measurable space because we are able to measure it if we add some structure by defining a measure $\mu: \mathcal F \to \overline{\mathbb R}_+$ (to be compared with a measure space $(X, \mathcal F, \sigma)$ where we can measure everything (ish)). A measurable function is then a function $f: X \to Y$ that preserves the measurable property of the spaces on which it is defined.