Prove Riemann Integral is Zero

$$\mathbf{Theorem}: \$$If $$S\subset\mathbb{R}^N$$ is a non-empty measurable set, with Jordan measure (content) zero, and $$f:S\mapsto \mathbb{R}^M$$ is a bounded function, then the Riemann integral, $$\int_Sf=0$$

$$\mathbf{Lemma \ 1}:$$ A set $$S\subset \mathbb{R}^N$$ has Jordan measure zero, if for each $$\epsilon>0$$, there are compact intervals $$I_1,...,I_n\subset \mathbb{R}^N$$ with,

$$S\subset \bigcup_{i=1}^nI_i \ \ and \ \ \sum_{i=1}^n\mu(I_j)<\epsilon_0$$

Where $$\mu(I)$$ denotes the Jordan measure of the set $$I$$.

$$\mathbf{Proof:}$$

Suppose $$S\subset \mathbb{R}^N$$ is a non-empty measurable set, with Jordan measure zero, and $$f:S\mapsto \mathbb{R}^M$$ is a bounded function.

By $$\mathbf{Lemma \ 1}$$, a set $$S\subset \mathbb{R}^N$$ has Jordan measure zero, if for each $$\epsilon>0$$, there exists compact intervals $$\{I_i\}_{I=1}^N$$, such that $$S\subset \bigcup_{I=1}^NI_i$$ and $$\sum_{I=1}^N\mu(I_i)<\epsilon_0$$.

Then for $$\epsilon_0=\frac {\epsilon}{sup(f)}$$, clearly,

$$\bigg|\int_Sf\bigg|\leq\bigg|\int_{\bigcup_{I=1}^NI_i}f \bigg|\leq|sup(f)|\sum_{i=1}^N\mu(I_i)<\epsilon$$

That is, $$\int_Sf=0$$.

Is this a sufficient proof for the above theorem, or am I doing something wrong? Can anyone provide any feedback, or a proper proof if this isn't right? Thanks in advance!

I assume you want to prove this using the tools available for Riemann integration.

First, it has not been established that $f$ is Riemann integrable on $S$ since this is not given as a hypothesis. This can be shown as part of the correct proof. Second, the bounding argument is invalid for functions that change sign.

Since $S$ has zero content, it is bounded and is contained inside a bounded rectangle $Q$. The Riemann integral of $f$ over $S$ is defined as

$$\int_Sf = \int_Q g,$$

where $g(x) = f(x)$ for $x \in S$ and $g(x) = 0$ for $x \in Q \setminus S$.

Note that $g$ is Riemann integrable, since it is continuous everywhere except possibly at points of $S$ and the content of $S$ is zero.

Now take any partition $P$ of $Q$. Any subrectangle $R$ of $P$ has non-zero content (by definition of a partition). Hence $R$ is not a subset of $S$ and must contain at least one point where $g(x) = 0$. This implies that $\inf_R g(x) \leqslant 0 \leqslant \sup_R g(x)$. Forming upper and lower Riemann sums we have for any partition $P$

$$L(P,g) \leqslant 0 \leqslant U(P,g).$$

Hence,

$$\underline{\int}_Q g = \sup_P L(P,g) \leqslant 0 \leqslant \inf_PU(P,g) = \overline{\int}_Q g.$$

Since $G$ is Riemann integrable it follows that

$$\int_Sf = \int_Q g = \underline{\int}_Q g = \overline{\int}_Q g = 0.$$