A function which conserves null-sets

Question

Let $F: \Bbb{R} \hookrightarrow \Bbb{R}$ a continuous map that satisfies $\forall x < y : f(x) < f(y)$ and \begin{equation} \tag{I} \label{I} \lambda(A) = 0 \rightarrow \lambda(F(A))=0 \end{equation} where $\lambda$ is the Lebesgue measure. If $G(x)=x+F(x)$, then also $G$ satisfies \ref{I}.

I am getting stuck with this. If I am not wrong, it can be proved using the definition. I mean, given $\varepsilon > 0$ there exists $\{I_k=(a_k, b_k)\}_{k\in \Bbb{N}} : A \subseteq \bigcup_{k\in \Bbb{N}} I_k \land \sum_{k\in \Bbb{N}} l(I_k) < \varepsilon$. The problem is that I do not know how to use the increasity of $F$ to relate this with a cover for $F(A)$ and then for $G(A)$.

Thanks in advance :)

Answer 1

If $I=(a,b)$ is an interval, then $$\lambda(G(I))=(b+F(b))-(a+F(a))=(b-a)+(F(b)-F(a))=\lambda(I)+\lambda(F(I)).$$

By considering disjoint unions of open intervals, we can establish the same result for open sets: $$\lambda(G(U))=\lambda(U)+\lambda(F(U)).$$

Suppose that $\epsilon>0$ and $\lambda(A)=0$. Then we can find an open set $U\supseteq A$ with $\lambda(U)<\epsilon/2$ Further, since $\lambda(F(A))=0$, we can find an open set $V\supseteq F(A)$ such $\lambda(V)<\epsilon/2$. Since $F$ is continuous, $F^{-1}(V)$ is open, hence $F^{-1}(V)\cap U$ is too, and

$$\lambda(G(F^{-1}(V)\cap U))=\lambda(F(F^{-1}(V)\cap U))+\lambda(F^{-1}(V)\cap U)\leq \lambda(V)+\lambda(U)=\epsilon.$$

Answer 2

Let $A \subseteq \mathbb{R}$ be compact and null. We first show that $\lambda(G(A)) = 0$. Let $\epsilon > 0$. Then, as $A$ is null, there exists open intervals $\{I_k\}_{k \in \mathbb{N}}$ s.t. $A \subseteq \cup_k I_k$ and $\sum_k l(I_k) < \frac{\epsilon}{2}$. As $A$ is compact, there is a finite subcover, i.e., there exists finitely many open intervals $\{I_j\}_{j = 1}^n$ with $A \subseteq \cup_{j = 1}^n I_j$ and $\sum_{j = 1}^n l(I_j) < \frac{\epsilon}{2}$. By adjoining two open intervals together if they intersect, we may assume $I_j$ are disjoint. We may further assume $I_j$ are ordered in accordance with the usual order on the real line. Let $I_j = (a_j, b_j)$. As $F$ is strictly increasing, we have $F(\cup_{j = 1}^n I_j) = \cup_{j = 1}^n (F(a_j), F(b_j))$. We observe that these intervals are disjoint and also ordered in accordance with the usual order on the real line.

Now, as $F$ is continuous, $F(A)$ is compact. Per our assumptions, $F(A)$ is also null. So by the same argument we see that there exists finitely many disjoint open intervals $\{L_k\}_{k = 1}^m$ with $F(A) \subseteq \cup_{k = 1}^m L_k$ and $\sum_{k = 1}^m l(L_k) < \frac{\epsilon}{2}$. Furthermore, we shall assume $L_k$ are ordered in accordance with the usual order on the real line. Since $A \subseteq \cup_{j = 1}^n I_j$ and $F(\cup_{j = 1}^n I_j) = \cup_{j = 1}^n (F(a_j), F(b_j))$, we also have $F(A) \subseteq \cup_{j = 1}^n (F(a_j), F(b_j))$. We may delete all $j$ such that $(F(a_j), F(b_j))$ does not intersect any $L_k$. Indeed, if $x \in I_j$ for some $j$ s.t. $(F(a_j), F(b_j))$ does not intersect any $L_k$, then $F(x) \in (F(a_j), F(b_j)) \notin \cup_k L_k \supseteq A$, so $F(x) \notin F(A)$ and $x \notin A$. Therefore, deleting all such $j$ does not affect the covering of $A$. We may therefore assume WLOG that every $(F(a_j), F(b_j))$ intersects some $L_k$.

Now, both $\{L_k\}_{k = 1}^m$ and $\{(F(a_j), F(b_j))\}_{j = 1}^n$ are disjoint, ordered collections of open intervals, so for each $j$ there exists $k^1_j, k^2_j \in \{1, \cdots, m\}$ s.t. $(F(a_j), F(b_j))$ intersects $L_k$ iff $k^1_j \leq k \leq k^2_j$ and furthermore we have $k^1_1 \leq k^2_1 \leq k^1_2 \leq k^2_2 \leq \cdots \leq k^1_n \leq k^2_n$. Let $L'_{jk} = (F(a_j), F(b_j)) \cap L_k$ when $k^1_j \leq k \leq k^2_j$. Then $F(A) \subseteq \cup_{j = 1}^n \cup_{k = k^1_j}^{k^2_j} L'_{jk}$. $L'_{jk}$ are disjoint nonempty open intervals which, when ordered lexicographically in $j$ then $k$, are ordered in accordance with the usual order on the real line. As such, for each $j$ there exists $a_j \leq c^1_{j, k^1_j} < c^2_{j, k^1_j} \leq c^1_{j, k^1_j + 1} < c^2_{j, k^1_j + 1} \leq \cdots \leq c^1_{jk^2_j} < c^2_{jk^2_j}$ s.t. $F(c^1_{j, k})$ (resp., $F(c^2_{j, k})$) is the left (resp., right) endpoint of $L'_{jk}$ for all $k^1_j \leq k \leq k^2_j$. Then, for any $x \in A \cap I_j$, there must exists a unique $k^1_j \leq k \leq k^2_j$ s.t. $x \in (c^1_{j, k}, c^2_{j, k})$. So $G(x) = x + F(x) \in (c^1_{j, k} + F(c^1_{j, k}), c^2_{j, k} + F(c^2_{j, k}))$. Hence,

$$G(A) \subseteq \cup_{j = 1}^n \cup_{k = k^1_j}^{k^2_j} (c^1_{j, k} + F(c^1_{j, k}), c^2_{j, k} + F(c^2_{j, k}))$$

So $\lambda(G(A)) \leq \sum_{j = 1}^n \sum_{k = k^1_j}^{k^2_j} [(c^2_{j, k} - c^1_{j, k}) + (F(c^2_{j, k}) - F(c^1_{j, k}))]$. We have,

$$\sum_{j = 1}^n \sum_{k = k^1_j}^{k^2_j} (c^2_{j, k} - c^1_{j, k}) \leq \sum_{j = 1}^n l(I_j) < \frac{\epsilon}{2}$$

And,

$$\sum_{j = 1}^n \sum_{k = k^1_j}^{k^2_j} (F(c^2_{j, k}) - F(c^1_{j, k})) = \sum_{j = 1}^n \sum_{k = k^1_j}^{k^2_j} l(L'_{jk}) \leq \sum_{k = 1}^m l(L_k) < \frac{\epsilon}{2}$$

Thus, $\lambda(G(A)) \leq \epsilon$. As $\epsilon$ is arbitrary, this shows $\lambda(G(A)) = 0$ whenever $A$ is compact and null.

The general case follows from inner regularity. Indeed, if $A$ is null, to show $G(A)$ is null, it suffices to show all compact subsets $K \subseteq G(A)$ is null. $G$ is strictly increasing, continuous, and clearly surjective, so it admits a continuous inverse $G^{-1}$. Thus, $G^{-1}(K)$ is a compact subset of $A$, whence null. So $K = G(G^{-1}(K))$ is null as well.