To show $\Lambda_f(R)$ is finite if and only if $f$ is bounded, where $f$ is monotone increasing Let  $f$ is monotone increasing
To show $\Lambda_f(R)$ is finite if and only if f is bounded, where $\Lambda_f$ is Lebesgue-Stieltjes outer measure
($\Leftarrow$) I have proved it.
But how to approach ($\Rightarrow$) ? Since $f$ is not right continuous so we can't use $\Lambda_f((a,b])=f(b)-f(a)$
 A: If $f$ is right continuous , then for such interval $(a,b]$ we have $$\Lambda_f(a,b]=f(b)-f(a)$$
And consider the special interval, we have $$\Lambda_f(-n,n]=f(n)-f(-n) $$
And the both side direction will be done by passing $n\rightarrow\infty$ and use the continuity of outer measure.
Now, if the problem does not assume $f$ is right continuous. Then there exists a counterexample for ($\Rightarrow$)
Define $f(x)=0$ for $x\leq 1$ and $f(x)=n$ if $x\in (n,n+1]$, where $n\in N$
Consider $R=\displaystyle\bigcup_{n\in N}(n,n+1]$, then since the union is disjoint $$\Lambda_f(R)=\displaystyle\sum_{n\in Z} \Lambda_f (n,n+1]$$
Observe that $$ \Lambda_f (n,n+1] =0$$ for $n\in Z$ and $n<1$
Hence, we focus on $n\geq 1$
Consider $(\alpha, n+1]\subseteq (n,n+1]$
Then $$0\leq \Lambda_f(\alpha,n+1]\leq f(n+1)-f(\alpha)=0$$. Since $f$ is constant on$(n,n+1]$
This is true for any $\alpha$ with $n<\alpha\leq <n+1$. Hence,
$$\Lambda_f(n,n+1] =\displaystyle\lim_{\alpha\rightarrow n} \Lambda_f(\alpha,n+1] =0$$
Hence, $$\Lambda_f(R)=\displaystyle\sum_{n\in Z} \Lambda_f (n,n+1] =0$$
But $f$ is unbounded.
