$F(x)=\sum_{n}^{\infty}\frac{1}{2^n}\chi_{[r_n, \infty)}$ is right continuous

Question

Let $\left\{ r_n\right\}$ be enumeration of rational numbers. Prove
(1) F is continuous from the right
(2) F is continuous at $x$ if $x$ is irrational
$$F(x)=\sum_{n}^{\infty}\frac{1}{2^n}\chi_{[r_n, \infty)}$$

I have shown that $F$ is strictly increasing ($x < y \Rightarrow F(x)<F(y)$) and $F$ is bounded (in particular $F(-\infty)=0, \ F(+\infty)=1$).

I thought if I consider a sequence $(x_n)\searrow x$, then $F(x)< \cdots <F(x_2)<F(x_1)$ and monotone boundness tells us $F(x_n) \rightarrow F(x)=\inf_n\left\{ F(x_n)\right\}$, but I'm not sure about this.

Answer 1

You can directly use the definition of continuity.

Fix $x_0 \in \mathbb{R}$, we show $f$ is right-continuous at $x_0$. Given $\varepsilon > 0$, choose $\delta$ so small such that all $\{r_n\}$ in the interval $(x_0,x_0+\delta)$ satisfies $2^{-n} < \varepsilon$. When $x\in (x_0,x_0+\delta)$, we have $$|F(x) - F(x_0) | = \sum_{n: r_n \in (x_0,x)} \frac{1}{2^n} \leq \sum_{n: r_n \in (x_0,x_0+\delta)} \frac{1}{2^n} \leq \varepsilon \sum_{n=0}^\infty \frac{1}{2^n} = 2\varepsilon $$ this established the right-continuity.

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Proving $f$ is continuous at an irrational $x_0$ is similar. Given $\varepsilon > 0$, choose $\delta$ so small such that all $\{r_n\}$ in the interval $(x_0-\delta,x_0+\delta)$ satisfies $2^{-n} < \varepsilon$, we can do this because $x_0$ is irrational. Then proceed as above.

Answer 2

For each $x\in\mathbb{R}$, denote the Dirac-measure associated with $x$ by $\delta_x$. That is, for any Borel set $A\subseteq\mathbb{R}$, $\delta_x(A) = 1$ if $x\in A$ and $\delta_x(A)=0$ if $x\not\in A$. Define $\mu = \sum_{n=1}^\infty \frac{1}{2^n} \delta_{r_n}$, which is a Borel measure. Then $F$ defined at the beginning is the cumulative distribution function associated with $\mu$. That is, $\mu( (-\infty, x]) = F(x)$. The right-continuity of $F$ follows from the general theory about the c.d.f. of a Borel measure.

To show that $F$ is the c.d.f. of $\mu$, let $x\in\mathbb{R}$. Then $\mu((-\infty, x]) = \sum_{n=1}^\infty \frac{1}{2^n} \delta_{r_n}( (-\infty,x])$. For each $n$, $\delta_{r_n}( (-\infty,x]) =1$ iff $r_n \in (-\infty,x]$ iff $r_n \leq x$ iff $x\in [r_n,\infty)$. Therefore $\chi_{[r_n,\infty)}(x)=\delta_{r_n}( (-\infty,x])$ and it follows that $\mu((-\infty, x]) = F(x)$.

To show that $F$ is continuous at $x$ if $x$ is irrational: Let $x\in\mathbb{R}$ be an irrational number. Let $(x_n)$ be an aribrary sequence of real numbers such that $x_1<x_2<\ldots<x$ and $x_n\rightarrow x$. To show that $F$ is left-continuous at $x$, it suffices that $\lim_n F(x_n)=F(x)$ (Heine Theorem). Note that $F(x)-F(x_n) = \mu( (x_n,x] ) = \mu( (x_n, x) )$ because $\mu(\{x\})=0$ (as $x$ is irrational). Note that $\{(x_n,x) \mid n\in\mathbb{N}\}$ is a decreasing sequence of Borel sets with $\bigcap_n (x_n,x) = \emptyset$. By continuity of measure, we have $\lim_n \mu( (x_n,x)) = \mu(\bigcap_n (x_n,x)) =0$. That is, $\lim_n F(x_n)=F(x)$.