Let $f:X \rightarrow (Y, \mathcal{Y})$ be an abstract function, with $\mathcal{Y}$ a $\sigma$-algebra on $Y$. Endow $X$ with $f^{-1}(\mathcal{Y})$. Is then $f(X)$ a measurable set in $Y$? If not, are there simple conditions on $f$ making $f(X)$ measurable? If $\mathcal{Y}$ were a $\sigma$-ring, would this modify anything?
More concisely (and generally): when is the image of a measurable set under a measurable function a measurable set?
Best Answer
There are actually positive results if you change the context a little bit. Suppose that $X$ is a separable complete metric space, i.e., a Polish space, and assume that $Y$ is something like $\mathbb R^n$, a Polish space that carries a measure that interacts nicely with the topology like the Lebesgue measure.
Now, if $f:X\to Y$ is Borel measurable, then for every Borel set $B\subseteq X$ the image $f[B]$ is not necessarily Borel in $Y$, but it is Lebesgue measurable in $Y$.