I'm looking for a proof of this statement
Let $E ⊆ ℝ^n$ be a set of Jordan content zero (it means that for each $ε > 0$ there are compact rectangles $E_1,…,E_m$ such that $E ⊆ ∪_jE_j$ and $∑_jvol(E_j) < ε$). If $I = [a_1,b_1]×…×[a_n,b_n]$ ($a_i < b_i$) and $f : I → ℝ$ is bounded on $I$ and continuous on $I – E$, then $f$ is Riemann integrable on $I$.
I know the proof of the version in which f is supposed continuous, but here the presence of the zero content set complicates the matters.
Best Answer
The proof is not difficult, but the notation can become heavy when dealing with multidimensional intervals. I'm assuming you know about partitions of intervals, Darboux sums etc. If not you can find a complete treatment of Riemann n-dimensional integral in Munkres book "Analysis on manifolds".
For each $ε > 0$ exist $E_1,…,E_m ⊆ ℝ^n$, with $E_j = [a_1^j,b_1^j]×…×[a_n^j,b_n^j]$, such that $∑_{j=1}^m vol(E_j) < ε$ and $f$ is continuous on $I - D$, with $D = ∪_{j=1}^m E_j$.
Notice that you can suppose that the $E_j$s are non degenerate subrectangles of $I$. In fact if $E_j$ and $I$ are compact rectangles then $E_j∩I$ is also a compact rectangle and $vol(E_j∩I) ≤ min\{vol(E_j),vol(I)\}$. Beside if $I$ is non degenerate (i.e. $a_i < b_i$) and $E_j ⊆ I$ is degenerate, we can find a non degenerate rectangle $E_j'$, of arbitrarily small volume, such that $E_j ⊆ E_j' ⊆ I$.
Also, without loss of generality, you can assume f continuous on $C = I – D° = I - ∪_{j=1}^m(a_1^j,b_1^j)×…×(a_n^j,b_n^j)$. In fact, if $0 < h < min\{(b_i^j - a_i^j)/2 \ | \ 1 ≤ i ≤ n, 1 ≤ j ≤ m\}$ and $F_j = [a_1^j-h,b_1^j+h]×…×[a_n^j-h,b_n^j+h]$, then $vol(F_j) < 2^nvol(E_j) ⇒ ∑_{j=1}^m vol(Fj) < 2^n∑_{j=1}^m vol(E_j) < 2^nε$ with $ε$ arbitrarily small and $f$ continuous on $I - ∪_{j=1}^m E_j ⊇ I - ∪_{j=1}^mF_j°$.
Now, $C$ is compact, so, by Heine-Cantor theorem, we can find $δ > 0$ such that $x, x' ∈ C$ and $∥x'-x∥ < δ$ $⇒ |f(x')-f(x)| < ε$. Let's suppose $|f| ≤ K$, and let $P = (P_1,…,P_n)$ be a (multidimensional) partition of $I$ such that $mesh(P) < δ$. If $P_i^* = P_i ∪ (∪_{j=1}^m\{a_i^j,b_i^j\})$, $P^* = (P_1^*,…,P_n^*)$ and $J$ is the generic subrectangle of $I$ determined by $P^*$, then:
$$U(f,P^*) - L(f,P^*) = ∑_J osc(f,J)⋅vol(J) = ∑_{J⊆C}osc(f,J)⋅vol(J) + ∑_{J⊆D}osc(f,J)⋅vol(J) ≤ ∑_{J⊆C}ε⋅vol(J) + ∑_{J⊆D}2K⋅vol(J) ≤ ε⋅vol(I) + 2K⋅(∑_{J⊆E_1 }vol(J) +…+ ∑_{J⊆E_m}vol(J)) = ε⋅vol(I) + 2K⋅∑_{j=1}^m vol(E_j) < ε⋅[vol(I)+2K]$$
Because $ε$ is arbitrarily small, this inequality proves that $f$ is integrable on $I$.
As a side note, keep in mind that this result is not reversible, there are integrable function which are not continuous with the exception of a zero-content set. But it will become reversible if we switch from Peano-Jordan to Lebesgue measure. In fact there is a theorem (due to Lebesgue) which affirms that a function is (Riemann) integrable if and only if is continuous (Lebesgue) almost everywhere. You can find the (non trivial) proof of this result in Munkres book I mentioned earlier.