Let $I\subset\mathbb R^n$ be a $n-$ rectangle and $f,\,g:\ I\longrightarrow\mathbb R$ be bounded function such that $f=g$ over $I\setminus C$ where $C$ is of (Lebesgue) measure zero, which is defined as $$\forall\epsilon>0,\ \exists n-\text{rectangles } \{U_i\}_1^\infty:\ C\subset\bigcup_1^\infty U_i\ \text{ and }\ \sum_1^\infty|U_i|<\epsilon. $$ We know that a function is Riemannian integrable iff it is bounded and continuos almost everywhere. Thus $f-g$ is integrable. It's clearly that if $C$ is of (Jordan) content zero, then $$\intop_I(f-g)=0. $$ But what if $C$ is of measure zero ?
Here is my attempt. Let
$$T=I\setminus\left(\bigcup_1^\infty U_i \right), $$
then $T\cap C=\varnothing$, thus $f-g=0$ over $T$. Now, we choose a finite collection $\{U_i\}_1^m$ from $\{U_i\}_1^\infty$. Let $P$ be a partition of $I$ by taking all endpoints of $\{U_i\}_1^m$. Then
\begin{align}
\sum_{\substack{R\in H(P) \\ R\subseteq T}}\left(\sup_R(f-g)\right)|R|=\sum_{\substack{R\in H(P) \\ R\subseteq T}}\left(\inf_R(f-g)\right)|R|=0
\end{align}
If $R\not\subseteq T$ then $R$ is contained in some $U_i$. Hence, with $M\geq|f-g|$, we have
\begin{align}
\sum_{\substack{R\in H(P) \\ R\not\subseteq T}}\left(\sup_R(f-g)\right)|R|\leq\sum_{\substack{R\in H(P) \\ R\not\subseteq T}}M|R|\leq\sum_{1}^mM|U_i|\leq \sum_{1}^\infty M|U_i|< M\varepsilon
\end{align}
Similarly,
$$\sum_{\substack{R\in H(P) \\ R\not\subseteq T}}\left(\inf_R(f-g)\right)|R|>-M\varepsilon. $$
Therefore,
$$-M\varepsilon<L(f-g,P)\leq\intop_I(f-g)\leq U(f-g,P)<M\varepsilon $$
Since $\varepsilon$ is arbitrary, this implies
$$\intop_I(f-g)=0\ \text{ or }\ \intop_If=\intop_Ig. $$
Is my proof valid ? Is there any nice other way to solve this ? Thanks
Assuming $f$ and $g$ are Riemann integrable and equal almost everywhere, it follows that $\int_I f = \int_I g$.
Let $P$ be any partition of rectangle $I$ with subrectangles $\{R_j\}_{j=1}^n$, where, by definition, each subrectangle is non-empty and has positive measure.
Since $C$ has measure zero, we have $R_j \not\subset C$ for each $j$. This implies that each subrectangle contains a point $x$ such that $h(x) := f(x) - g(x) =0$, and, hence,
$$\inf_{x \in R_j} h(x) \leqslant 0 \leqslant \sup_{x \in R_j} h(x),$$ Forming lower and upper sums it then follws that
$$L(P,h) \leqslant 0 \leqslant U(P,h)$$
Since $P$ was arbitrary this implies that
$$\underline{\int_I}h=\sup_{P}L(P,h) \leqslant 0 \leqslant \inf_{P}U(P,h)= \overline{\int_I}h$$
But $h = f-g$ is Riemann integrable and the lower and upper integrals are equal, Whence,
$$\int_I h = \underline{\int_I}h= \overline{\int_I}h=0$$