$f: \mathbf{R} \rightarrow \mathbf{R}$ monotone increasing $\Rightarrow$ $f$ is measurable

If $f$ is increasing, the set {$x:f(x)>a$} is an interval for all $a$, hence measurable. By definition (Royden's), the function $f$ is measurable.

It is true that $f$ is continuous almost everywhere, but it is not true that there exists a continuous function $g:\mathbb R\to\mathbb R$ such that $f=g$ almost everywhere, unless $f$ is already continuous, as Jacob said in a comment. Note that the left-hand and right-hand limits of $f$ exist everywhere, and if they are not equal for $f$, then they cannot be equal for any function equal to $f$ almost everywhere. E.g., the characteristic function of $[0,\infty)$ is monotone and not equal almost everywhere to a continuous function.

(The characteristic function of the rationals is equal almost everywhere to a continuous function, but is continuous nowhere. This shows from the other direction why "continuous almost everywhere" and "equal almost everywhere to a continuous function" are very different.)

Modifying your work to something correct takes more effort than the simpler and clearer solution given by Cass. Let $D$ be the set of discontinuities of $f$. Then $D$ is countable, hence of measure $0$. The restriction $f|_D$ is measurable on $D$ because every subset of $D$ is measurable, and the restriction $f|_{\mathbf R\setminus D}$ is measurable on $\mathbf R\setminus D$ because it is continuous. You can show this implies that $f$ is measurable.