Problem 3

Exercise 9.3 (Measuring with an expanding ruler)   Let $A\subseteq \mathbb{R}$ be a set of positive finite measure. Define a function

$$\varphi : \mathbb{R} \to [0,\infty)$$ (9.12)

$$x \mapsto \mu(A\cap(-\infty,x])$$ (9.13)

1. Show that $\varphi$ is continuous.
2. Find $x\in\mathbb{R}$ such that $\mu(A\cap(-\infty,x))=\mu(A\cap(x,\infty))$.

Proof. Let us show that $\varphi$ is continuous. Let $\epsilon>0$ be given and $x\in\mathbb{R}$. Let $0<\vert x-y\vert<\epsilon$. We show $\vert\varphi(x)-\varphi(y)\vert<\epsilon$ also in two similar cases. Let $y>x$. Then $(-\infty,x]\subseteq(-\infty,y]$, so we know that

$$\varphi(y)-\varphi(x) = \mu(A\cap(-\infty,y])-\mu(A\cap(-\infty,x]) = \mu(A\cap(-\infty,y]\setminus A\cap(-\infty,x]) = \mu(A\cap(x,y])$$ (9.14)

$$\leq \mu((x,y]) \leq \epsilon$$ (9.15)

For $y< x$, make a similar argument.

To prove the existence of $x\in\mathbb{R}$ such that $\mu(A\cap(-\infty,x))=\mu(A\cap(x,\infty))$, examine the difference

$$d(x) = \mu(A\cap(-\infty,x))-\mu(A\cap(x,\infty))$$ (9.16)

By the continuity of measure and $\varphi$, this can be written

$$d(x) = \varphi(x) - (\mu(A)-\varphi(x))=2\varphi(x) - \mu(A)$$ (9.17)

One can plainly see the limits

$$\lim_{x\to\infty} \varphi(x) = \mu(A)$$ (9.18)

$$\lim_{x\to-\infty} \varphi(x) = 0$$ (9.19)

Therefore, $d(-\infty)=-\mu(A)$ and $d(\infty)=\mu(A)$, so the Intermediate Value Theorem summons $x\in(-\infty,\infty)$ such that $d(x)=0$ and $\mu(A\cap(-\infty,x))=\mu(A\cap(x,\infty))$. $\qedsymbol$