If $g$ is bounded and satisfies $g(x) = f(x)$, then $g$ is integrable and $\int_{a}^{b} g(x)dx = \int_{a}^{b} f(x)dx$.

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Let $f$ be integrable on $[a,b]$ and $E$ be a finite subset of $[a,b]$.

Show that if $g$ is bounded function which satisfies $g(x) = f(x)$, $\forall x\in[a,b]\backslash E$, then $g$ is integrable on $[a,b]$ and $\int_{a}^{b} g(x)dx = \int_{a}^{b} f(x)dx$.

Proof: Suppose $g$ is bounded function which satisfies $g(x) = f(x)$, $\forall x\in[a,b]\backslash E$, Then since $g(x)$ is bounded so is the difference of $f(x) - g(x)$. Then let $h(x) = f(x) - g(x) = 0$ for some finite numbers of [a,b] . Then$\int_{a}^{b} h(x)dx = \int_{a}^{b} [f(x) - g(x)]dx = \int_{a}^{b} f(x)dx -\int_{a}^{b} g(x)dx = 0$.

Then $\int_{a}^{b} g(x)dx = \int_{a}^{b} f(x)dx$.

Is my proof ok? Any feedback/help would be really appreciated. Thank you.