= Solution
Every <Lebesgue measurable set> $A$ differs from a <Borel set> $B$ by a <Lebesgue measure>[Lebesgue-null] set $N$. The given function $h$ is a <homeomorphism> from $\mathbb R$ onto the interval $h(\mathbb R)$. Hence $h(B)$ is Borel relative to $h(\mathbb R)$ and therefore Lebesgue measurable in $\mathbb R$. Moreover,
$$
h(A)\mathbin{\triangle}h(B)\subseteq h(A\mathbin{\triangle}B)\subseteq h(N),
$$
and $h(N)$ is null by <Lusin condition N>. The <Completeness of Lebesgue measure> now shows that $h(A)$ is measurable, so
$$
\nu(A):=\lambda(h(A))
$$
is defined for every $A\in\mathcal M$.
Because $h$ is <injective function>[injective], it maps pairwise disjoint sets to pairwise disjoint sets and commutes with arbitrary unions. The <countable additivity> of $\lambda$ therefore gives
$$
\nu\left(\bigcup_{j=1}^{\infty}A_j\right)
=\lambda\left(\bigcup_{j=1}^{\infty}h(A_j)\right)
=\sum_{j=1}^{\infty}\lambda(h(A_j))
=\sum_{j=1}^{\infty}\nu(A_j).
$$
Thus $\nu$ is the <image-length measure of a strictly increasing continuous function>. If $\lambda(A)=0$, the null-set hypothesis gives $\nu(A)=\lambda(h(A))=0$, so
$$
\boxed{\nu\ll\lambda}.
$$
The measure is <sigma-finite measure>[sigma-finite], since
$$
\nu([-m,m])=h(m)-h(-m)<\infty.
$$
The <Radon-Nikodym theorem> consequently supplies a nonnegative locally integrable function $f$ such that
$$
\nu(A)=\int_Af\,d\lambda.
$$
For $a<x$, strict increase and continuity give
$$
h([a,x])=[h(a),h(x)]
$$
and hence
$$
h(x)-h(a)=\nu([a,x])=\int_a^x f(t)\,dt.
$$
The analogous oriented identity holds for $x<a$. By <differentiation of an indefinite Lebesgue integral>,
$$
\boxed{h'(x)=f(x)\quad\text{for almost every }x}.
$$
\b[Yes, the differentiability conclusion still holds when $h$ is merely non-decreasing.] In fact, the <Lebesgue theorem on differentiability of monotone functions> says that every <monotone function> is differentiable almost everywhere; neither continuity nor the null-set hypothesis is needed for that conclusion. Flat intervals mean that the preceding image-set argument no longer gives disjoint images, so its natural replacement is the <Lebesgue-Stieltjes measure> determined by
$$
\boxed{\mu_h((a,b])=h(b)-h(a).}
$$
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