If $A \subset R^d$ is measurable with finite lebesgue measure, and I have the function: $f(y) = m(A \cap \{x: x_1 < y \})$ then it's clear that this is a continuous function of $y$ and also that as $y \to \infty$, $f(y) \to m(A)$, and as $y \to -\infty$, $f(y) \to 0$.
This is obvious to me as the set $\{x: x_1 < y \}$ contains nothing as we make $y$ very small and negative and vice versa.
But I am wondering how to rigorously show this?
In order to prove continuity let $(y_n)_n$ be a convergent sequence in $\mathbb R^d$ with $y=\lim_{n\to\infty}y_n$.
Then it is enough to prove that: $$\lim_{n\to\infty} f(y_n)=\lim f(y)\tag1$$
Observe that $f$ is monotonically increasing.
For that reason it is enough to prove $(1)$ under the extra condition that the sequence $(y_n)_n$ is monotonically increasing and secondly under the extra condition that the sequence is monotonically decreasing.
Setting $B_n=\{x\in\mathbb R^d\mid x_1<y_n\}$ and $B=\{x\in\mathbb R^d\mid x_1<y\}$ in the first case we are dealing with $B_1\subseteq B_2\subseteq B_3\subseteq\cdots\subseteq B=\bigcup_{n=1}^{\infty}B_n$ and find: $$\lim_{n\to\infty}f(y_n)=\lim_{n\to\infty}m(B_n)=m(\bigcup_{n=1}^{\infty}B_n)=m(B)=f(y)$$
In the second case we are dealing with $B_1\supseteq B_2\supseteq B_3\supseteq\cdots\supseteq\bigcap_{n=1}^{\infty}B_n=B\cup C$ where $C$ is a measurable subset of $\{x\in\mathbb R^d\mid x_1=y\}$.
Then the facts that $m(B_1)<\infty$ and $m(\{x\in\mathbb R^d\mid x_1=y\})=0$ allow the conclusion:$$\lim_{n\to\infty}f(y_n)=\lim_{n\to\infty}m(B_n)=m(\bigcap_{n=1}^{\infty}B_n)=m(B\cup C)=m(B)=f(y)$$
In sortlike way (ascending sets and descending sets) it can be shown that $\lim_{y\to\infty}f(y)=m(A)$ and $\lim_{y\to-\infty}f(y)=m(\varnothing)=0$.