I'm confused about the proof below from Donald Cohn's Measure Theory:
In particular I don't see how compactness of $[t,x]$ is used in the end of the argument.
The following seems to work, but does not require any of the $\delta_{n}$'s.
Once $t<x$ is chosen in the second paragraph so that $F(t) < \epsilon$, what is wrong with the following simpler argument?
$\begin{eqnarray*}F(x) - \epsilon &<& F(x) - F(t)\\ &=& \mu^{*}((t,x])\text{, (from the check that }\mu^{*}\text{ is an outer measure)}\\ &\leq& \mu^{*}(-\infty,x]\text{, (by monotonicity)} \end{eqnarray*}$
yielding $$F(x) \leq \mu^{*}(-\infty,x] + \epsilon$$.
Sorry for this oversight. The problem is that $\mu^{*}[t,x]$ need not be $F(x) - F(t)$, since $F$ is only continuous from the right.