We know that absolutely continuous functions $g\in {\rm W}^{1,1}(0,1)$ have the Lusin N-property: if $E\subseteq (0,1)$ is a set of measure zero, then $g(E)$ is also a set of measure zero. My question is do absolutely continuous functions allow stronger properties, and if they do not, what smaller class of functions allows such properties? Here are extensions of Luzin N-property I am interested in: Consider $g\in {\rm W}^{1,1}(0,1)\cap {\rm C}^1(0,1)\cap {\rm C}[0,1]$. Which of the following assertions is true, and for which class of functions (for instance, if one of the assertions is not true for absolutely continuous functions, it could be true for locally Lipschitz functions):

  1. If $F\subseteq (0,1)$ is a set of full measure in $(0,1)$, then $g(F)$ is a set of full measure in ${\rm Im}(g)$ (Remark: here ${\rm Im}(g):=\{g(s):s\in [0,1]\}$, and so, by uniform continuity of $g$, it holds that ${\rm Im}(g)=[m,M]$ for some $m,M\in\mathbf{R}$ such that $m\leq M$)

  2. For every sequence of measurable subsets $(E_n)$ in $(0,1)$ such that $\lim_{n\rightarrow+\infty}\lambda(E_n)=0$ it holds that $\lim_{n\rightarrow+\infty}\lambda(g(E_n))=0$.

  3. For every $\varepsilon>0$ there exists $\delta>0$ such that: for every countable family of non-overlapping intervals $([a_k,b_k])_{k=1}^{+\infty}$ such that $\sum_{k=1}^{+\infty}|b_k-a_k|\leq\delta$ we have $\sum_{k=1}^{+\infty}|g(b_k)-g(a_k)|\leq\varepsilon$. Remark. We know that a weaker property, namely when choice is restricted to finitely many intervals as above provides that $g$ is (locally) absolutely continuous. If we allow finitely many overlapping intervals, then $g$ is locally Lipschitz, i.e., $g\in {\rm W}^{1,\infty}_{loc}(0,1)$; these remarks can be found in G. Leoni's book: A First Course in Sobolev Spaces.

  4. For every $\varepsilon>0$ there exists $\delta>0$ such that: for every countable family of (possibly overlapping) intervals $([a_k,b_k])_{k=1}^{+\infty}$ such that $\sum_{k=1}^{+\infty}|b_k-a_k|\leq\delta$ we have $\sum_{k=1}^{+\infty}|g(b_k)-g(a_k)|\leq\varepsilon$.

  5. For every $\varepsilon>0$ there exists $\delta>0$ such that: for every measurable set $E\subset (0,1)$ such that $\lambda(E)\leq\delta$ we have $\lambda(g(E))\leq\varepsilon$. Remark. We know that $\lambda(E)\leq\delta$ implies $\int_{E}|g(s)|ds\leq\varepsilon$, this is so-called equi-integrability property of $g\in {\rm W}^{1,1}(0,1)$.

  6. If $(K_m)$ is an increasing sequence of compact sets in $(0,1)$ such that $\lim_{m\rightarrow+\infty}\lambda((0,1)\backslash K_m)=0$, then $\lim_{m\rightarrow+\infty}\lambda\{\xi\in\mathbf{R}: (g_{\vert K_m})^{\leftarrow}(\xi)\neq\emptyset\}=\lambda({\rm Im}(g))$, where, as before, ${\rm Im} (g):=g([0,1])=[m,M]$.

  7. Measure preserving mapping: for every $0<\theta<1$ there exists $0<\varphi_{\theta}<1$ such that we have: for every measurable $E\subseteq (0,1)$ such that $\lambda(E)=\theta$, it holds that $\lambda(g(E))=\varphi_{\theta}\lambda({\rm Im}(g))$.

I tried to sort the assertions in an descending order so as to introduce stronger and stronger requirements on $g$. So, I guess, property 1. should be easy to prove for absolutely continuous functions, and it is not to restrictive, while property 7. is rather restrictive and probably holds only for (a sub-class of) locally Lipschitz functions (property 7. holds, for instance, if $g$ is an affine function).


2 Answers 2


So far, I found the answer to 4., The definition of absolute continuity without disjointness of intervals is satisfied only by Lipschitz functions

whereby it follows that 4. is satisfied provided $g$ is Lipschitz function;

And I also found the answer to 3., Absolutely Continuous (is the collection of subintervals supposed to be finite or infinite)

whereby it follows that 3. is equivalent to choosing finitely many non-overlapping intervals, so that absolutely continuous functions satisfy 3.


The following is for 5.

First lemma: Every absolutely continuous function is the difference between two monotone and absolutely continuous functions.

If F is absolutely continuous then f is an indefinite integral, so F(x)= $\int_{a}^{x}f(t)dt $ and therefore, F(x)=$\int_{a}^{x}f^{+}(t)dt $ $-$ $\int_{a}^{x}f^{-}(t)dt $.

These two functions are absolutely continuous beacuse they are indefinite integrals and monotone because both $f^{+}$ and $f^{-}$ are $\geq$0.

Now with this let's proceed to:

Second lemma: If A is such that $m^{*}(A)$=a, then for every $\epsilon$, there is a sequence ($I_{n}$) of disjoint open intervals s.t. A $\subset \bigcup_{n\geq1} I_{n}$ and $\sum l(I_{n}) \leq a+\epsilon.$

By definition of outer measure, we can find a sequence ($I_{n}$) of open intervals s.t. A $\subset \bigcup_{n\geq1} I_{n}$ and $\sum l(I_{n}) \leq a+\epsilon.$ (though it doesn't need to be disjoint)

As $\bigcup_{n\geq1} I_{n}$ is open, it can be rewritten as a union of disjoint open intervals $J_{n}$!


$A \subset \bigcup_{n\geq1} I_{n}=\bigcup_{n\geq1} J_{n} $ and $ \sum l(J_{n}) = (*_{1}) = m(\bigcup_{n\geq1} J_{n})= m(\bigcup_{n\geq1} I_{n}) \leq \sum l(I_{n}) \leq a+\epsilon$

[$(*_{1})$ is true because they are disjoint!]

So if we take open intervals, we can assume they all are disjoint always :)

Third lemma: If f is absolutely continuous and increasing, then it satisfies $\forall \epsilon$ $ \exists \delta$: $m^{*}(A) \leq \delta \implies m^{*}(f(A)) \leq \epsilon$.

$m^{*}(A) \leq \delta$ so there is a sequence ($I_{n}$) of disjoint open intervals s.t. A $\subset \bigcup_{n\geq1} I_{n}$ and $\sum l(I_{n}) \leq 2\delta.$

We have $f(A) \subset \bigcup_{n\geq1} f(I_{n})$ and $f(I_{n})$ are always intervals if f is continuous.

In this !special case!, as !f is increasing!, $f((a,b)) =(f(a),f(b))$.

($I_{n}$) of disjoint open intervals and $\sum l(I_{n}) \leq 2\delta$ and because f is abolutely continuous, then $\sum l(f(I_{n})) \leq \epsilon$

Therefore, $(f(I_{n}))$ is a sequence of open intervals, $f(A) \subset \bigcup_{n\geq1} f(I_{n})$ and $\sum l(f(I_{n})) \leq \epsilon$, so we conclude $m^{*}(f(A)) \leq \epsilon$

Now, for the finale!

h absolutely continuous, so $h=f-g$, with $f$ and $g$ increasing, absolutely continuous.

Suppose $m^{*}(A) \leq \delta$. Again, so there is a sequence ($I_{n}$) of disjoint open intervals s.t. A $\subset \bigcup_{n\geq1} I_{n}$ and $\sum l(I_{n}) \leq 2\delta.$

$h(A) \subset \bigcup_{n\geq1} h(I_{n})$

A minor detail is that $\forall x \in [a,b]: h(x)=f(x)-g(x) \leq f(b)-g(x) \leq f(b)-g(a)$. So $|h(x)-h(y)| \leq f(b)-g(a)+g(b)-f(a) = f(b)-f(a) + g(b)-g(a)$.

So if $I_n=[a_n,b_n]$, $l(h(I_n)) \leq f(b_n)-f(a_n) + g(b_n)-g(a_n) = l(f(I_n))+l(g(I_n))$.

We know from before, if ($I_{n}$) is sequence of disjoint open intervals s.t. $\sum l(I_{n}) \leq 2\delta$, then $\sum l(f(I_{n})) \leq \epsilon$ and $\sum l(g(I_{n})) \leq \epsilon$.

So $l(h(I_n)) \leq l(f(I_n))+l(g(I_n)) \leq 2\epsilon$.

$h(A) \subset \bigcup h(I_{n})$, $h(I_{n})$ are intervals and $l(h(I_n)) \leq 2\epsilon$.

So we can conclude that $m^{*}(h(A)) \leq 2\epsilon$.


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