Constructing a measure space given the definition of a measurable function

Question

Suppose:

• $(X,\mathcal{S},\mu)$ is a (complete) measure space,
• $(\mathbb{R},\mathcal{B}_\mathbb{R})$ is the Borel space associated to $\mathbb{R}$, and
• $f:(X,\mathcal{S})\to(\mathbb{R},\mathcal{B}_\mathbb{R})$ is a measurable function.

Given this information, is it possible to construct the measure space $(Y,\mathcal{T},\lambda)$ defined by:

• $Y=f(X):=\{f(x):x\in X\}\subset\mathbb{R}$,
• $\mathcal{T}=f(\mathcal{S}):=\{f(A):A\in\mathcal{S}\}\subset\mathcal{B}_\mathbb{R}$,
• $\lambda=f_*(\mu)$ (i.e. the pushforward of $\mu$ by $f$).

If possible, do we need to consider the completion of $(X,\mathcal{S},\mu)$ for the above statement to hold?

Thank you, Frederick

Answer 1

In the general case, you can not expect that $f(\mathcal{S}):=\{f(A):A\in\mathcal{S}\}\subseteq\mathcal{B}_\mathbb{R}$.

Here is a simple counter-example:

Consider $X=\mathbb{R}$, $\mathcal{S}=2^\mathbb{R}$ and $\mu$ the counting measure. Then $(X,\mathcal{S},\mu)$ is a complete measure space. Consider the function $f:(X,\mathcal{S})\to(\mathbb{R},\mathcal{B}_\mathbb{R})$ defined by $f(x)=x$. It is easy to see that $f$ is a measurable function. Now, take any $E\subseteq \mathbb{R}$ such that $E \notin \mathcal{B}_\mathbb{R}$. Then $E\in 2^\mathbb{R}$ and $f(E)=E \notin \mathcal{B}_\mathbb{R}$. So $f(\mathcal{S})\nsubseteq\mathcal{B}_\mathbb{R}$.

So to do what you want to do, you need to define the $\sigma$-algebra in different way. Let us see it.

Let us construct the measure space $(Y,\mathcal{T},\lambda)$ defined by:

• $Y=f(X):=\{f(x):x\in X\}\subset\mathbb{R}$,
• $\mathcal{T}:=\{A \cap Y:A\in\mathcal{B}_\mathbb{R}\}$,
• $\lambda=f_*(\mu)$ (i.e. the pushforward of $\mu$ by $f$).

Note that, for any $E\in \mathcal{T}$ there is $A\in\mathcal{B}_\mathbb{R}$ such that $E=A \cap Y$ and $\lambda(E) = \mu(f^{-1}(A))$.

Note that $\lambda$ is well defined. In fact, given $E\in \mathcal{T}$ and $A_1, A_2\in\mathcal{B}_\mathbb{R}$ such that $E=A_1 \cap Y=A_2 \cap Y$, then $f^{-1}(A_1)=f^{-1}(A_2)$ and so $\mu(f^{-1}(A_1))=\mu(f^{-1}(A_2))$.