Real Analysis Folland Problem 1.5.28 Borel Measures 
Problem 1.5.28 - Let $F$ be increasing and right continuous, and let $\mu_F$ be the associated measure. Then $\mu_F(\{a\}) = F(a) - F(a-)$, $\mu_F([a,b)) = F(b-) - F(a-)$,$\mu_F([a,b]) = F(b) - F(a-)$, and $\mu_F((a,b)) = F(b-)- F(a)$

Proof - $\mu_F$ has been constructed on the algebra of h-intervals that takes the values $$\mu_F((a,b]) = F(b) - F(a) \forall a,b$$ So $\mu_F$ is a finite bounded set on $\mathbb{R}$ We can represent $\{a\}$ as $$\{a\} = \bigcap_{n=1}^{\infty}(a - 1/n, a]$$ The interval $(a - 1/n, a]$ forms a decreasing sequence of sets so by theorem 1.8d we have $$\mu_F(\{a\}) = \lim_{n\rightarrow \infty}\mu_F((a - 1/n, a]) = \lim_{n\rightarrow \infty}\left[ F(a) - F(a - 1/n)\right]$$ Since $F$ is increasing $$\lim_{n\rightarrow \infty}\mu_F(a - 1/n) = \lim_{x\rightarrow a-}F(x) = \sup\left\lbrace F(x):x < a\right\rbrace  = F(a-)$$ Thus $$\mu_F(\{a\}) = F(a) - F(a-)$$ Now, consider the $\mu_F([a,b))$, where $b$ might be $\infty$. We can represent $[a,b) = \{a\} \cup (a,b)$ (disjoint union) so $$mu_F([a,b)) = \mu_F(\{a\}) + \mu_F(a,b) = F(a) - F(a-) + F(b-) + F(a) = F(b-) - F(a-)$$ Note this is valid if $F(b-) = F(\infty) = \infty$ Next, consider $\mu_F([a,b])$ We can represent $[a,b] = \{a\} \cup (a,b]$ (disjoint union) so $$mu_F([a,b]) = \mu_F(\{a\}) + \mu_F([a,b]) = F(a) - F(a-) + F(b) - F(a) = F(b) - F(a-)$$ Finally, consider $\mu_F((a,b))$ so either $a$ or $b$ will be $\infty$ in the interval. If $b$ is $\infty$ then $(a,\infty)$ is an h-interval and the definition of $\mu_F$ yields $$\mu_F((a,\infty)) = F(\infty) - F(a)$$ so the proposed formula $$\mu_F((a,b)) = F(b-)- F(a)$$ is correct because $F(\infty) = \lim_{x\rightarrow \infty}F(x)$
The part I don't understand is why $$\lim_{n\rightarrow \infty}F(a-1/n) = F(a-)$$
Try to explain this as simply as possible.
 A: Your reasoning is in good direction.

Problem 1.5.28 - Let $F$ be increasing and right continuous, and let $\mu_F$ be the associated measure. Then $\mu_F(\{a\}) = F(a) - F(a-)$, $\mu_F([a,b)) = F(b-) - F(a-)$,$\mu_F([a,b]) = F(b) - F(a-)$, and $\mu_F((a,b)) = F(b-)- F(a)$

Proof: 
Part 1. $\mu_F$ has been constructed on the algebra of h-intervals that takes the values $$\mu_F((a,b]) = F(b) - F(a) \:\:\:\forall a,b\in \mathbb{R}$$ We can represent $\{a\}$ as $$\{a\} = \bigcap_{n=1}^{\infty}(a - 1/n, a]$$ The intervals $(a - 1/n, a]$ form a decreasing sequence of sets and $\mu_F((a-1,a]) <+\infty$, so by theorem 1.8d we have 
$$\mu_F(\{a\}) = \lim_{n\rightarrow \infty}\mu_F((a - 1/n, a]) = \lim_{n\rightarrow \infty}\left[ F(a) - F(a - 1/n)\right]=F(a) -\lim_{n\rightarrow \infty}F(a - 1/n)  \tag{1}$$ 
Since $F$ is increasing, we know that $\lim_{\substack{x\to a \\ x<a} }F(x)$ exists (and is finite), and it is easy to see that 
$$\lim_{n\rightarrow \infty}F(a - 1/n)=\lim_{\substack{x\to a \\ x<a} }F(x)$$
 but since $F$ is assumed to be just right continuous, we can NOT conclude $\lim_{\substack{x\to a \\ x<a} }F(x)=F(a)$. The notation $F(a-)$ is used exactly to mean $\lim_{\substack{x\to a \\ x<a} }F(x)$. So, from$(1)$, we have  
$$\mu_F(\{a\}) = F(a) -\lim_{n\rightarrow \infty}F(a - 1/n) = F(a)-F(a-)$$ 
Note that for all $a\in \mathbb{R}$, $\mu_F(\{a\}) <+\infty$
Part 2. Given $a, b \in \mathbb{R}$ we have $[a,b)=\{a\} \cup ((a,b] \setminus \{b\})$.
So 
\begin{align*} 
\mu([a,b))&= \mu(\{a\})+ \mu((a,b] - \{b\}) =\mu(\{a\})+ \mu((a,b]) - \mu(\{b\}) = \\
& = F(a)-F(a-) +F(b)-F(a) -(F(b)-F(b-)) =\\
& = F(b-)-F(a-)
\end{align*}
Part 3. Given $a, b \in \mathbb{R}$ we have $[a,b]=\{a\} \cup (a,b] $.
So 
\begin{align*} 
\mu([a,b])&= \mu(\{a\})+ \mu((a,b]) = \\
& = F(a)-F(a-) +F(b)-F(a) =\\
& = F(b)-F(a-)
\end{align*}
Part 4. Given $a, b \in \mathbb{R}$ we have $(a,b)=(a,b] \setminus \{b\}$.
So 
\begin{align*} 
\mu((a,b))&= \mu((a,b] - \{b\}) = \mu((a,b]) - \mu(\{b\}) = \\
& = F(b)-F(a) -(F(b)-F(b-)) =\\
& = F(b-)-F(a)
\end{align*}
Remark: For the cases where $a=-\infty$ or $b=+\infty$ (or both), we have:


*

*If $a=-\infty$ and $b \in \mathbb{R}$, then  using the family $\{(-n, b]\}_n$  and  $\{(-n, b)\}_n$,
$$\mu_F((-\infty,b])=F(b)-F(-\infty+)$$
and 
$$\mu_F((-\infty,b))=F(b-)-F(-\infty+)$$

*If $a  \in \mathbb{R}$ and $b =+\infty$, then  using the family $\{(a, n]\}_n$  and  $\{[a, n]\}_n$,
$$\mu_F((a,+\infty))=F(+\infty-)-F(a)$$
and 
$$\mu_F([a,+\infty))=F(+\infty-)-F(a-)$$

*If $a =-\infty$ and  $b =+\infty$, then  using the family $\{(-n, n]\}_n$ ,
$$\mu_F((-\infty,+\infty))=F(+\infty-)-F(-\infty+)$$
Note in the those three cases, we may have $F(+\infty-)=+\infty$ or $F(-\infty+)=-\infty$ (or both).
