a basic doubt on continuous image of a measurable set measurable Is continuous image (say the function is on $\Bbb R^n$) of a measurable set measurable ? Hint enough. Actually, $f: \Bbb R^n \to \Bbb R^n$ is given to be linear. I used some theorem to conclude that it is continuous.
 A: Here is an answer in the case that the set is Lebesgue measurable and $f$ is Lipschitz.
Suppose $f$ is Lipschitz with rank $L$, and use the $\max$ norm for convenience. Let $S$ be an open 'square', that is, $S=\Pi_k I$ for a bounded interval $I=(a,b)$. Since $S = \cup_i S_i$, where $S_i = \Pi_k [a+{1 \over i}, b-{1 \over i}] $, we have $f(S) = \cup_i f(S_i)$, and since each $S_i$ is compact, we have $f(S_i)$ is compact and hence measurable. It follows that $f(S)$ is measurable. Since we can write an open 'rectangle' $R$ as the countable union of squares, we see that $f(R)$ is measurable as well.
Then if we let $x=({a+b \over 2}, ..., {a+b \over 2})$, and $y \in S$, then $\|f(x)-f(y)\| \le L \|x-y\| < L ( {b-a \over 2})$
and so $f(S) \subset B(f(x), L({b-a \over 2}))$.
It follows that $m(f(S)) \le L^n m(S)$. Since we can write an open rectangle as the countable union of squares, we have $m(f(R)) \le L^n m(R)$ for any open rectangle $R$. It follows that $\cup_k R_k$ is also measurable and $m(f(\cup_k R_k)) \le \sum_k m(f (R_k)) \le L^n \sum_k m(R_k)$.
Key fact:
Suppose $f$ is Lipschitz and the set $N$ has Lebesgue measure zero. Then for any $n>0$ there is a collection of open rectangles $R_k$ such that $N \subset \cup_k R_k$ and $\sum_n m R_k < \epsilon$. Since $f(N) \subset f(\cup_k R_k)$, we see that $m^*(f(N)) < L^n \epsilon$, where $m^*$ is the outer measure. It follows that $m^*(f(N)) = 0$ and since the Lebesgue measure is complete, we have that $f(N)$ is measurable and $m(f(N))=0$.
Now suppose $A$ is measurable. Then there is a $F_\sigma$ set $F$ such that $F \subset A$ and $m(A \setminus F) = 0$ (note that from above, we have that $f(A \setminus F)$ is measurable). We can write $F = \cup_i F_i$, where each $F_i$ is compact. As above, since $f$ is continuous we  have $f(F_i)$ is compact, hence measurable. It follows that $f(F) = f(\cup_i F_i) = \cup_i f(F_i)$ is measurable, and since $A = F \cup (A \setminus F)$, we have $f(A) = f(F) \cup f(A \setminus F)$, the union of two measurable sets, hence $f(A)$ is measurable.
It follows from the above that $m(f(A)) \le L^n m(A)$.
Since any linear map on a finite dimensional space is Lipschitz, we see that a linear map maps measurable sets to measurable sets.
