# Show that $f\colon\bar{M}\to\mathbb{R}$ is Riemann-integrable

Consider a Jordan-measurable set $M\subset\mathbb{R}^n$ and $f\colon\bar{M}\to\mathbb{R}$ continious. Show that $f$ is Riemann-integrable over $M$.

I think the main facts for the prove are:

• $M$ is bounded and so $\bar{M}$ is compact

• $f$ therefore is bounded and uniformly continious

I need a stepfunction that is $\leqslant f$ and one that is $\geqslant f$.

In a book I found this idea of a proof:

Set $B:=\sup_{x\in M}\lvert f(x)\rvert$.

Consider any $\varepsilon >0$. Then there are finite many disjoint open intervalls $(Q_j)$ so that $$\mathrm{vol}(M)-\sum_j \mathrm{vol}(Q_j)\leqslant\varepsilon/(2B).$$ Because $f$ is uniformly continious one can divide each $Q_j$ small enough in intervals $Q_j'$ so that the oscillation on each $Q_j$ is $\leqslant \varepsilon/(2 \mathrm{vol}(M)$.

Equally there are disjoint compact intervalls $(K_l)$ so that $\mathrm{vol}(M)\geqslant \sum_l \mathrm{vol}(K_l)-\varepsilon/(2B)$ and that the oscillation on each $K_l$ is $\leqslant \varepsilon/(2 vol(M))$.

From this we get for each $x\in M$ $$\sum_j (\inf_{y\in Q_j}f(y))\cdot\chi_{Q_j}(x)\leqslant f(x)\leqslant\sum_l (\sup_{y\in K_l}f(y))\cdot\chi_{K_l}(x)$$ and from this $$\sum_j (\inf_{y\in Q_j}f(y))\cdot vol(Q_j)\leqslant\int_* f\leqslant\int^* f\leqslant\sum_l (\sup_{y\in K_l}f(y))vol(K_l)$$

Last but not least the book says that it is (by choice of the $Q_j$ and the $K_l$) $$\lvert\sum_l (\sup_{y\in K_l}f(y))vol(K_l)-\sum_j (\inf_{y\in Q_j}f(y))vol(Q_j)\rvert\leqslant\varepsilon.$$

There is one main thing I did not understand yet:

Why is $$\left\lvert\sum_l (\sup_{y\in K_l}f(y))vol(K_l)-\sum_j (\inf_{y\in Q_j}f(y))vol(Q_j)\right\rvert\leqslant\varepsilon?$$

With greetings

Revised answer: So it seems that the proof from your book is incorrect or incomplete. So please allow me to write a different proof from scratch, which is hopefully simpler:

Let a simple function be a finite sum of the form $$g(x) = \sum_i a_i \mathbb{1}_{A_i}(x)$$ with $a_i \in \mathbb{R}$, the $A_i$'s a disjoint union of Jordan measurable sets, and $\mathbb{1}_{A_i}$ the indicator function on $A_i$. Then we define the Riemann integral of $g$ to be $$\int g = \sum_i a_i |A_i|$$ where $|\cdot|$ is the Jordan measure of the set.

Let the lower sum $\int_{*,M} f$ on $M$ be defined to be $$\int_{*,M} f = \sup \int g$$ where the supremum is taken over all simple functions $g:M \rightarrow \mathbb{R}$ with $g \leq f$ on $M$. We similarly define the upper sum $\int_M^* f$. If $\int_{*,M} f = \int_M^* f$, we define their common value to be the Riemann integral $\int_M f$ of $f$ over $M$.

Now let $Q_i \subset \mathbb{R}^n$ be a disjoint collection of half-open $n$-rectangles that cover $M$, such that the oscillation of $f$ on each $Q_i \cap M$ is less than $\epsilon / M$.

Let $A_i = Q_i \cap M$. Note that since intersections of Jordan measurable sets are Jordan measurable, the $A_i$ are Jordan measurable. Then $$g(x) = \sum_i (\inf_{A_i} f) \mathbb{1}_{A_i}(x)$$ is a simple function on $M$ with $g \leq f$. Similarly let $$h(x) = \sum_i (\sup_{A_i} f) \mathbb{1}_{A_i}(x)$$ be the simple function on $M$ with $f \leq h$.

Then we have $$\int_M g \leq \int_{*,M} f \leq \int_M^* f \leq \int_M h.$$ Moreover, $$\int_M h - \int_M g = \sum_i (\sup_{A_i} f - \inf_{A_i} f) |A_i| \leq \sum_i \frac{\epsilon}{M} |A_i| = \epsilon.$$

Since $\epsilon$ can be made arbitrarily small, we have $\int_{*,M} f = \int_M^* f$, showing the $f$ is Riemann integrable on $M$.

• Concerning point 1): We defined: $\int_* f:=\sup\left\{\int\tau | \tau\in\mathbb{T}_c(\mathbb{R}^N, t\leqslant f\right\}\in\bar{\mathbb{R}}$, where $\mathbb{T}_c(\mathbb{R}^N)$ denotes the space of the N-dimensional step functions. Is that the definition you mean? – math12 May 22 '14 at 10:23
• Another question: Isn't your first "$\leqslant$" a "$=$"? And why does from $\ldots\leqslant\frac{3}{2}\varepsilon + \frac{1}{2MB}\varepsilon^2$ follow that the term is $\leqslant\varepsilon$? – math12 May 22 '14 at 10:25
• Where in the proof is standing that the $K_l$ are not disjoint from $M$? – math12 May 22 '14 at 12:01
• Moreover it is not clear to me, if $Q:=\bigcup_j Q_j\subset M$ why then it is $\sum_j (\inf_{y\in Q_j} f(y))\chi_{Q_j}(x)\leqslant f(x)$ for $x\in M$? Because what is if $x\in M\setminus Q$? Then the step function 0, but why is $f(x)\geqslant 0$ then? – math12 May 22 '14 at 12:24
• It is true that it would work only for $f \geq 0$. But that problem is in your proof as well since the $Q_j$ are not specified to cover $M$. So the proof from the book is either incorrect or incomplete. So rather than patch together this proof, I'll rewrite a proof of your result from scratch. If you also want someone to check your book's proof, you should give a specific reference. – John M May 22 '14 at 16:14