Image of a continuous mapping and measureability issues Suppose $f : \mathbb{R} \rightarrow \mathbb{R}$ is a continuous mapping.
Then, if I understand correctly:


*

*The image of $f$ needn't be Lebesgue measurable.

*The pushforward of the Lebesgue measure across $f$, being a Borel measure, has well-defined support, and this support is a closed set.

*Hence the image of $f$ needn't agree with the support of the pushforward of the Lebesgue measure.



Questions.
Q0. Is this correct? If so, an instructive example or two would be appreciated.
Q1. Would it be fair to the say that the support of the pushforward of the Lebesgue measure across $f$ is always the closure of the image of $f$? If not, what is the correct statement of the relationship between the image of $f$ and the support of the pushforward of the Lebesgue measure?

 A: Q0: no, not correct.  The image of $f$ can be written
$$f(\mathbb{R}) = \bigcup_n f([-n,n])$$
which is a countable union of compact (hence closed) sets.  So it is Borel (even $F_\sigma$) and Lebesgue measurable.
This still happens if you replace $\mathbb{R}$ by another Polish space; the continuous image of a Polish space into another Polish space is (by definition) analytic, and all analytic sets in a Polish space are universally measurable (though not necessarily Borel).  The same happens if you only require $f$ to be Borel measurable.
But it is true that the image of $f$ need not equal the support of the pushforward, since the support must be closed and the image need not be.  Example: $f(x) = \arctan x$.
Q1: let $\mu$ be the pushforward of $m$ under $f$ and let $E$ be the support of $\mu$.  The inclusion $E \subset \overline{f(\mathbb{R})}$ is an immediate consequence of definitions.  Conversely, let $U = E^c$.  Then $f^{-1}(U)$ is an open set of Lebesgue measure zero, so it must be empty.  That means $f(\mathbb{R}) \subset E$.  Since $E$ is closed we also have $\overline{f(\mathbb{R})}\subset E$.
