Prove or disprove $\nu(E)=\lambda(f(E))$ is a measure provided that $f$ is nondecreasing and satisfies the N-condition. Suppose $f$ is a non-decreasing continuous function from $[a,b]$ to $\mathbb{R}$, and $\lambda$ is the Lebesgue measure in $\mathbb{R^1}$. Also, $f$ satisfies the property that $f$ maps Lebesgue measurable set to Lebesgue measurable set. Prove or disprove $\nu(E)=\lambda(f(E))$ is a measure.
Based on the hypothesis above, if $f$ also maps zero measurable set to zero measurable set, prove or disprove $\nu(E)=\lambda(f(E))$ is a measure.
For the first question, I don't believe $\nu$ is a measure, but I don't have a good example at hand. For the second question, I learned from somewhere that $f$ is actually absolutely continuous, thus $\nu$ has to be a measure. But I don't know how to prove this directly without proving $f$ is absolutely continous. Any ideas would be appreciated.
 A: We first note that a continuous function $f : [a,b] \to \Bbb{R}$ maps Lebesgue measurable sets to Lebesgue measurable sets iff it maps null-sets to null-sets.
For one implication note that (by regularity) every measurable set $M \subset [a,b]$ can be expressed as a $\sigma$-compact set plus a null-set, i.e. $M = N \cup \bigcup_n K_n$ with $K_n$ compact and $N$ a null-set. Hence,
$$
f(M) = f(N) \cup \bigcup_n f(K_n)
$$
is measurable, since $f(N)$ is a null-set by assumption and $f(K_n)$ is compact and hence measurable.
For the converse, assume that there is a null-set $N \subset [a,b]$ such that $f(N)$ is not a null-set, i.e. of positive measure. It is known (see Positive outer measure set and nonmeasurable subset) that every set of positive (outer) measure contains nonmeasurable sets. Hence, there is a nonmeasurable $V \subset f(N)$. But since $V$ is contained in $f(N)$, we have
$$
V = f(f^{-1}(V) \cap N),
$$
where $f^{-1}(V) \cap N$ is a subset of a null-set and hence measurable. By assumption, this yields that $V$ is measurable as the image of a measurable set under $f$, contradiction.

Next, let us show that $\nu$ is always a measure as long as $f(E)$ is measurable for all measurable $E$. This even holds without continuity assumptions on $f$.
First,  since $f$ is nondecreasing, we know that $I_y := f^{-1}(\{y\}) \subset [a,b]$ is an interval for all $y \in \Bbb{R}$. Since the $(I_y)_y$ are pairwise disjoint, there are at most countably many $(y_n)_n$ for which $I_{y_n}$ contains more than one point (why exactly?).
Note that $L := \bigcup_n I_{y_n}$ is measurable as a countable union of intervals.
Now $f$ is injective on $M := [a,b] \setminus L$ (which might be the empty set), because if there were $x,y \in M$ with $x<y$ but $z := f(x) = f(y)$, this would imply that $x,y \in I_z$ and hence $z = y_n$ for some $n$. But $x,y \in I_z = I_{y_n}$ in contradiction to $x,y \in M$.
Finally, we have
$$f(L) = \bigcup_n f(I_{y_n}) \subset \bigcup_n \{y_n\},$$
which is a countable set. But this implies
$$
\nu(E) = \lambda(f(E)) = \lambda(f(E \setminus L)) = \lambda(f(E \cap M)) =: \gamma(E)
$$
for each measurable $E \subset [a,b]$.
Using the fact that $f$ is injective on $M$, it is easy to see that $\gamma$ is a measure, since if $(E_n)_n$ is a pairwise disjoint sequence, then so is $(E_n \cap M)_n $ and thus also $(f(E_n \cap M))_n$, which implies
$$
\gamma(\bigcup_n E_n) = \lambda(\biguplus_n f(E_n \cap M)) = \sum_n \lambda(f(E_n \cap M)) = \sum_n \gamma(E_n).
$$
Hence, $\nu = \gamma$ is always a measure.

As observed at the beginning, we must always have that $f(E)$ is a null-set if $E$ is. Hence, $\nu \ll \lambda$, where $\lambda$ is Lebesgue measure.
By the Lebesgue-Radon-Nikodym theorem, we get $f \in L^1 ([a,b])$ with
$$
\nu(E) = \int_E f \, dt.
$$
In particular, if $f$ is continuous, then
$$
\lambda([f(x), f(y)]) = \nu([x,y]) = \int_x^y f(t) \, dt
$$
holds for $a \leq x < y \leq b$, so that $f$ is absolutely continuous.
