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$\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!

RRL
  • 90,707
The math god
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1 Answers1

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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

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

RRL
  • 90,707