If $f = 0$ in $I\subset \mathbb{R}^n$ almost everywhere, then $\int_I f = 0$ 
Suppose we have an $n-$dimensional interval $I \subset \mathbb{R}^n$ and a Riemann integrable function $f: I \to \mathbb{R}$ such that $f \neq 0$ only on a set $E \subset I$ of measure zero. Then $\displaystyle \int_I f =0. $

I would like a proof for this property without using only Measure theory.
I know that $E$ has measure zero in $\mathbb{R}^n$ if and only if for every $\varepsilon > 0$, there exists an at most countable covering consisting only of $n-$dimensional intervals $({I_n})_{n=1}^\infty$ of $E$ such that $$\sum_{n=1}^\infty |I_n| < \varepsilon,$$ where $|I_n|$ denotes the volume of the interval $I_n$.
But I cannot see how to use this property in order to show that every Riemann sum of $f$ is smaller than $\varepsilon$. 
I would have to prove that $\forall \varepsilon > 0, \exists \delta_\varepsilon >0$ such that for every partition $P= \{J_1, J_2, \cdots, J_m \}$ of $I$ with $\displaystyle \max_{J_i \in P}(|J_i|) < \delta_\varepsilon$, we have that $$|\sigma(f, P, \xi)| = \left| \sum_{i=1}^{n} f(x_i) \cdot |J_i| \right|  < \varepsilon, $$ where $\xi$ is a system of intermediate points in $P$ i.e. $\xi = \{\xi_1, \xi_2, \cdots, \xi_m \}$ with $\xi_i \in J_i ,\forall j \in \{1,2,\cdots,m \}$. However, I really don't know how to choose $\delta_\varepsilon$ so that I can use the fact that $E$ has measure zero.
 A: Let $B=\{x\in I:f(x)=0\}$ and $E=\{x\in I:f(x)\neq 0\}$. 
We have the disjoint union $I=B\cup E$ and $$\int_If=\int_Bf+\int_Ef=\int_Ef$$
I am going to compute $\int_Ef$ using Riemann sums.
You have that the sets $B_n=\{x:n<|f(x)|\leq n+1\}$ have also measure $0$ for all $n\geq 0$. So, given $\epsilon>0$. Hence You can cover the set $B_n$ by "intervals" $\{I_{n,m}\}_m$ such that $\sum |I_{n,m}|<\epsilon/2^n$. Compute the Riemann sum of considering the cover $\{I_{n,m}\}_{n,m}$. What do you get?
A: Here's a proof using the Darboux formulation of integration, as in Spivak's Calculus on manifolds/ Munkres' analysis on manifolds, rather than the direct Riemann sum formulation.
Let $P$ be an arbitrary partition for the rectangle $I$. I claim that for every subrectangle $S$ determined by the partition, $S$ is not contained in $E$. If this wasn't true, there would be a subrectangle $S$ such that $S \subseteq E$. Since $E$ has measure zero, so does $S$. However, non-trivial closed rectangles do not have measure zero, so we would get a contradiction. 
Note that $S \nsubseteq E$ implies that $0 \in \{f(x): x \in S \}$. Hence, we have following inequalities:
\begin{align}
m_S(f) := \inf \{f(x): x \in S \} \leq 0, \\
M_S(f) := \sup \{ f(x): x \in S \} \geq 0.
\end{align}
Hence, the upper and lower (Darboux) sums satisfy
\begin{align}
L(f,P) := \sum_{S}m_S(f) \cdot \text{vol}(S) \leq 0 \leq \sum_{S} M_S(f) \cdot \text{vol}(S) =: U(f,P)
\end{align}
Since this is true for every partition, we get:
\begin{align}
\sup_{P} L(f,P) \leq 0 \leq \inf_P U(f,P)
\end{align}
(the sup and inf being taken over all partitions of the rectangle $I$). However, since $f$ was assumed to be integrable, the sup and inf are equal,thus leading us to:
\begin{align}
\int_I f := \sup_P L(f,P) = \inf_P U(f,P) = 0.
\end{align}
