Define the function $F\mathop: \mathbb{R} \to [0,\infty)$ by $\displaystyle{F(x):= \mu \left((-\infty,x] \right)}$. Prove the following.

Question

Let $\mu$ be a finite Borel measure on $\mathbb{R}$ (that is, $\mu$ is a measure on ${\cal B}_{\mathbb{R}}$ and $\mu(\mathbb{R}) <\infty$). Define the function $F\mathop: \mathbb{R} \to [0,\infty)$ by $\displaystyle{F(x):= \mu \left((-\infty,x] \right)}$. Prove:

1. $F$ is increasing on $\mathbb{R}$.
2. $F$ is right-continuous.

3. $\displaystyle{\lim_{x \to -\infty} F(x) = 0}$ and $\displaystyle{\lim_{x \to \infty} F(x) = \mu (\mathbb{R})}$.

4. If $\mu(\{x\}) = 0$ for each $x \in \mathbb{R}$, then $F$ is continuous on $\mathbb{R}$.

My progress:

At the moment, I have only been able to prove (1). Here is my attempt:

Suppose that $x,y\in \mathbb{R}$ such that $x<y$. Then we have $$(-\infty,x] \subseteq (-\infty,y].$$ It follows from the Monotonicity of $\mu$ that $$F(x) = \mu((-\infty,x]) \leq \mu((-\infty,y]) = F(y).$$

Answer 1

If $\{x_n\}$ decreases to $x$ then $(-\infty,x_n]$ decreases to $(-\infty,x]$ so $\mu (-\infty,x_n] \to \mu (-\infty,x]$. This proves 2).

Since $(-\infty,x]$ increases to $\mathbb R$ as $x$increases to $\infty$ we get $\mu (-\infty,x]$ tends to $\mu (\mathbb R)$. Also $(-\infty,x]$ decreases to the empty set as $x$ de creases to $-\infty$ so $F(x) \to 0$ as $x \to -\infty$. This proves 3).

If $\{x_n\}$ increases (strictly) to $x$ then $(-\infty,x_n]$ increases to $(-\infty,x)$ so $\mu (-\infty,x_n] \to \mu (-\infty,x)$. Hence $\mu (-\infty,x_n]$ tends to $\mu (-\infty,x)$ which is $\mu (-\infty,x]- \mu \{x\}$. 4) follows easily from this.