Let $a$ and $b$ be any real numbers such that $a < b$, and let $S$ be a (nonempty) subset of the closed bounded interval $[a, b]$ such that $S$ has measure $0$. Now let $f \colon [a, b] \longrightarrow \mathbb{R}$ and $g \colon [a, b] \longrightarrow \mathbb{R}$ be any two functions such that$$f(x) = g(x) \qquad \mbox{ for all } x \in [a, b] \setminus S. \tag{1} $$
If both $f$ and $g$ \textit{are} integrable on $[a, b]$, then is$$\int_a^b f(x) dx = \int_a^b g(x) dx?$$
A rigorous but elementary answer is requested!
I know that a (real-valued) function defined and bounded on $[a, b]$ is (Riemann) integrable on $[a, b]$ if and only if it is continuous almost everywhere on $[a, b]$, that is, if and only if the set of discontinuities of that function in $[a, b]$ is a set of measure $0$.
And, any subset of a set (of real numbers) of measure $0$ is also a set of measure $0$.
Now let $D_f$ and $D_g$ denote, respectively, the set of discontinuities of $f$ and of $g$ in $[a, b]$,
If $D_f = D_g$, then $f$ is integrable if and only if $g$ is.
So we assume that $D_f \neq D_g$. Since $f(x) = g(x)$ for each $x \in [a, b] \setminus S$, we must have$$S \cap D_f \ \neq \ S \cap D_g.$$But since $S$ has measure $0$, so do both the sets $S \cap D_f$ and $S \cap D_g$. In this case either both $f$ and $g$ are integrable or both are not according as whether $D_f \setminus S = D_g \setminus S$ is a set of measure $0$ or not.
Am I right?