Induction of measure by a measurable function

Question

Suppose:

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

Then, it is clear that one can induce a measure, $\nu$, on $(\mathbb{R},\mathcal{B}_\mathbb{R})$ by pushing forward the measure $\mu$ by $f$ as follows: $$\nu:=f_*\mu=\mu\circ f^{-1}:\mathcal{B}_\mathbb{R}\to[0,\infty]\\ B\mapsto\nu(B)=\mu\circ f^{-1}(B),$$ yielding therefore the induced measure space $(\mathbb{R},\mathcal{B}_\mathbb{R},\nu)$.

My question is: if we had to map every element of $\Sigma$ by $f$ (symbolically defined here as $f(\Sigma):=\{f(A):A\in\Sigma\}$), then:

1. Do we get as a result of doing this a $\sigma$-algebra on $\mathbb{R}$?
2. If so, is this $f(\Sigma)\subseteq\mathcal{B}_\mathbb{R}$?

If not, what about if $X$ is a topological space and $\Sigma$ the associated Borel $\sigma$-algebra. Does point 1. or 2. still not hold?

Thank you, Frederick.

Answer 1

If $X$ and $Y$ are sets, and $\mathscr P(X)$ and $\mathscr P(Y)$ are the corresponding power sets, then the correspondence $$ B\in \mathscr P(Y) \mapsto f^{-1}(B) \in \mathscr P(X) $$ has much beter properties if compared to the correspondence $$ A\in \mathscr P(X) \mapsto f(A) \in \mathscr P(Y). $$ The former respects unions, intersections and complementation, while the latter only respects unions.

For that reason things often work much better when sets are moved backwards, as opposed to forward, through functions. For example, a function between topological spaces is continuois iff the inverse image of open sets is open. Open functions, namely functions that map open sets to open sets, have a much more limited use in Math if compared to continuous functions.

This said, and adopting the terminology in the question, $f(\Sigma )$ is not always a $\sigma $-algebra, since $\mathbb R$ is not necessarily in $f(\Sigma )$, unless $f$ is surjective. Moreover, it might happen that $f(A)\cap f(B)$ is not in $f(\Sigma )$ even if $A$ and $B$ are in $Σ$ (try to find an example to illustrate this).

The second question also has a negative answer. The reason is that there are continuous functions $f:[0,1]\to \mathbb R$, such as the Devil staircase, sending the Cantor set $C$ to a set of positive Lebesgue measure. If we take $\Sigma $ to be the set of all Lebesgue measurable sets of $[0,1]$, then every subset $A$ of $C$ is in $\Sigma $, but $f(A)$ cannot all be Borel measurable simply because there are $2^{2^{\aleph_0}}$ subsets of $C$ and only $2^{\aleph_0}$ Borel measurable subsets of $\mathbb R$.

Even if $X$ is a topological space and $\Sigma $ is its Borel $\sigma $-algebra, question (2) may have a negative answer. For a trivial example let $X$ be any non-Borel subset of $\mathbb R$, with its induced topology, and let $f$ be the inclusion of $X$ in $\mathbb R$. Obviously $X$ is Borel in itself, but $f(X)$ is not Borel in $\mathbb R$.

For another perspective I suggest reading about analytic sets, where emphasis is put on direct image of Borel sets.