Jtdylktuy

Let $Fcolon [a,b] to Bbb R$ be continuous and increasing. Then:

$F$ satisfies the Lusin (N) property if and only if it is absolutely continuous.

Here, Lusin (N) property means that for all $Nsubseteq [a,b]$ with $mathfrak{m}(N)=0$ we have $mathfrak{m}(F[N])=0$, where $mathfrak{m}$ of course denotes Lebesgue measure.

That absolute continuity implies the Lusin (N) property is not hard to check directly using the definitions. I also have an outline of proof for the direct implication, which I can understand. This question is about fixing my first attempt for said implication (I feel I was very close).

Assume by contradiction that $F$ is not absolutely continuous, and take $epsilon > 0$ to witness that. Then, for every $k>0$ there is a disjoint collection of open intervals $I_{k,1}, ldots, I_{k,n_k}subseteq [a,b]$ such that $$sum_{j=1}^{n_k} mathfrak{m}(I_{k,j})< frac{1}{k} qquad mbox{but}qquad sum_{j=1}^{n_k} mathfrak{m}(F[I_{k,j}]) geq epsilon.$$(To write the negation of absolute continuity as above I am using that $F$ is increasing to take intervals into intervals, and that $F$ -- hence $mu_F$ -- is continuous to disregard the endpoints of the $F[I_{k,j}]$ in the second bound). I want to somehow the take the limit of these collections of intervals. This is my motivation for defining $$N doteq bigcap_{k >0} bigcup_{j=1}^{n_k} I_{k,j}.$$It is clear that $mathfrak{m}(N)=0$. I want to use the second estimate to contradict the Lusin (N) property. I thought of using the Lusin (N) property to say that $mathfrak{m}(F[N]) = 0$, and use outer regularity of $mathfrak{m}$ to get an open set $Osupseteq F[N]$ with $mathfrak{m}(O) = mathfrak{m}(Osetminus F[N]) < epsilon$, and do something with the open set $F^{-1}[O]$.

Question: How can I conclude the proof from here? If not, is any of this salvageable?

You have to modify your choice of the intervals, because the property "absolute continuous" can fail in 'many points'. This makes possible that $bigcup_{j=1}^{n_k} I_{k,j}$ and $bigcup_{j=1}^{n_{l}} I_{k,j}$ are disjoint for some $k ne j$. Thus your construction can lead to $N = emptyset$! This is the reason why you cannot deduce a contradiction in your argument.

One major point is that we need to ensure that $f$ is not absolute continuous on $bigcup_{j=1}^{n_k} I_{k,j}$. This allows us to choice the next intervals (in the $k+1$-step of the induction argument) to be a subset of $bigcup_{j=1}^{n_{l}} I_{k,j}$ and thus
$$lambda(F(N)) = lambda (bigcap_{n=1}^infty bigcup_{j=1}^{n_k} F(I_{k,j}) = liminf_{n rightarrow infty} lambda( bigcup_{j=1}^{n_k} F(I_{k,j})) ge varepsilon.$$
We used in the first step that $f$ is injective. (This is true, because $f$ is increasing!)

Finally, let us justify that we can choose the next intervals as subsets of the previous one. If for some $delta >0$ and for all choices of disjoint intervals $I_1, ldots, I_n subset [a,b]$ with $lambda(I_n) <delta$ the function $f$ is absolute continious, then it is on $[a,b]$.

Just take $[a,b] = I_1 cup ldots I_n$ with $ delta/2 < lambda(I_n) < delta.$ Then there are $delta_l>0$ such that for any disjoint $J_1,ldots,J_m subset I_l$ with $sum_{i=1}^m lambda(J_i) < delta_n$ we have $sum_{i=1}^m lambda(f(J_i)) < varepsilon/(2n)$.

Let $delta' := min(delta_1,ldots,delta_n,delta/4)$. For any disjoint intervals with $H_1,ldots,H_m subset [a,b]$ with $sum_{k=1}^m lambda(H_k) < delta'$, we know that $H_k$ has non-trivial intersection with at most two sets $I_i$ and $I_j$. The intervals $H_i cap I_j$ have lenght less than $delta_i$. Thus $sum_{k=1}^m lambda(f(H_k cap I_j)) <varepsilon/(2n)$. All in all, we find
$$ sum_{k=1}^m lambda(f(H_k)) leq 2 sum_{l=1}^n sum_{k=1}^m lambda(f(H_k cap I_j)) < varepsilon.$$