# What can be said about the measure of the image of a subset of $\mathbb R^d$ of measure zero under a measurable function?

Suppose $E \subset \mathbb R^d$ has measure $0$ and $f: \mathbb R^d \longrightarrow \mathbb R$ is measurable. Does $f (E)$ necessarily have measure $0$?

I tried to find a counter-example though I failed.It is clear that countable subset will not work for otherwise the image of it is again countable in $\mathbb R$ and hence has measure $0$. So we need to find an uncountable set of measure $0$ which maps to some set in $\mathbb R$ of positive measure under some measurable function in order to find a counter-example.How to find it?

Suppose $d>1$. $f(x_1,\dots,x_d) := x_1$ is a measurable map $\mathbb R^d\to\mathbb R$. Furthermore, the set $$E := \{x\in \mathbb R^d \mid x_2 = x_3 = \dots =x_d = 0 \}$$ is null, but $f(E) = \mathbb R$.
You can take the Cantor set in $\Bbb R$ (measure zero) and map it onto the closed interval $[0,1]$ by a continuous increasing function. So, even continuous functions on $\Bbb R$ don't preserve measure zero.