Let $k>n$ be positive integers, and let $f:\mathbb R^n \to \mathbb R^k$ be a real-analytic map (i.e. every component of $f$ is a real-analytic function).
Suppose that we have a measurable subset $B \subseteq \mathbb R^k$ of measure zero, such that $f^{-1}(B)$ has positive measure in $\mathbb R^n$. Is it true that $\text{Image}(f) \subseteq B$?
Does anything change if we assume that $B$ is closed?
Here are some nice special cases:
If $B=\{ 0\}$, then the answer is positive: The zero set of a non-zero analytic function has measure zero.
Similarly, if $B$ is a hyperplane in $\mathbb R^k$, then we can compose $f$ with a linear functional which vanishes exactly on this hyperplane and use the previous observation.
Note that if we only know that $B$ is contained in a hyperplane $H$, the only thing we can conclude is that the image of $f$ is contained in $H$. As Holding Arthur's answer shows, we cannot in general conclude that image of $f$ is contained in $B$.
Are you missing something in your question? Let $f:\mathbb{R} \to \mathbb{R}^2: x \to (x,x)$. Then the set $B=\{(x,x):0\leq x\leq1\}$ have zero measure in $\mathbb{R}^2$, and $f^{-1}(\{(x,x):0\leq x\leq1\})=[0,1]$ has lebesgue measure $1$. But clearly the image of $f$ is NOT in $B$. Please tell me if I have misread your question.