Exercise 1.1.23 (Bounded continuous functions are Riemann integrable)

Question

Let $f:[a,b]	o\mathbb{R}$ be a bounded, piecewise continuous function. Show that $f$ is Riemann integrable.

Elu Thingol · Accepted Answer

We proceed by showing several related results which together constitute for a proof for the above assertion.

\begin{lemma}
	Let $I$ be a closed interval, and let $f:I \to \mathbb{R}$ be a continuous function. Then $f$ is Riemann integrable.
\end{lemma}
\begin{proof}
Since $f$ is continuous on a compact interval $I=[a,b]$, it must be uniformly continuous. That is,
$$\forall \epsilon > 0 \quad \exists \delta > 0 \quad \forall x,y \in I: \quad |x-y| \leq \delta \implies |f(x) - f(y)| \leq \epsilon$$
Thus, fix an $\epsilon >0$. We can then find a $\delta > 0$ for $\epsilon/(b-a)$ for the uniform continuity. By Archimedean principle choose a corresponding $\frac{1}{n} < \delta$. Set $\Delta := m(I)/n = (b-a)/n$, and consider the partition $I_1 = [a,a+\Delta),\ldots,I_{n-1}=[a + (n-2) \Delta, a + (n-1)\Delta),I_n = [a+(n-1)\Delta, a + n\Delta]$. A key observation is that the supremum and the infimum of each interval $I_i$ do not differ by more than $\epsilon$. To see why, recall that from uniform continuity on $[a,b]$ follows the uniform continuity on each $I_i$. In other words, we have $f(x) - f(y) \leq \epsilon$ for all $x,y \in I_i$. Taking supremum over $x$ we obtain $\sup f(I_i) - f(y) \leq \epsilon$ for all $y \in I_i$. Taking infimum over $y$ we get $\sup f(I_i) - \inf f(I_i) \leq \epsilon$. Define a piecewise continuous functions $\overline{g} = \sum  \sup f(I_i)1_{I_i}, \underline{g} = \sum \inf f(I_i)1_{I_i}$; these are obviously piecewise continuous functions that majorize/minorize $f$. Then,
\begin{align*}
\text{p.c} \int_{[a,b]} \overline{g} - \text{p.c} \int_{[a,b]} \underline{g}
&= \sum_{i=1}^{n} \sup f([a + (i-1)\Delta, a + i\Delta)) \cdot m(I) - \sum_{i=1}^{n} \inf f([a + (i-1)\Delta, a + i\Delta)) \cdot m(I)\[5pt]
&= \sum_{i=1}^{n} \left[ \sup f([a + (i-1)\Delta, a + i\Delta))- \inf f([a + (i-1)\Delta, a + i\Delta)) \right] \cdot m(I)\
&\leq \sum_{i=1}^{n} \frac{\epsilon}{b-a} \cdot \frac{b-a}{n} = \epsilon
\end{align*}
Thus, $f$ must be Riemann integrable by Darboux criteria.
\end{proof}

\begin{lemma}
	Let I be an open interval. Let $f:I\to \mathbb{R}$ be a bounded continuous function. Then $f$ is Riemann integrable.
\end{lemma}
\begin{proof}
The idea is to partition $I=(a,b)$ into three intervals $(a,a+\epsilon),[a+\epsilon,b-\epsilon],(b-\epsilon,b)$. We have seen in the previous part that $f$ is Riemann integrable on $[a+\epsilon,b-\epsilon]$. Thus, we only need to demonstrate that $f$ is Riemann integrable on $(a,a+\epsilon)$ and $(b-\epsilon,b)$.      Fix an $\epsilon > 0$. Let $M$ be a bound of $f$, and consider the intervals $(a,a+\frac{\epsilon}{M})$ and $(b-\frac{\epsilon}{M},b)$. We then see that the upper and lower cover of $f$ on these intervals not only converge to each other, but vanishes altogether! Set $\overline{g} = \sup f(a,a+\frac{\epsilon}{M}), \underline{g} = \inf f(a,a+\frac{\epsilon}{M})$. Then      $$ \text{p.c} \int_{[a,b]} \overline{g} = \sup f(\left(a,a+\frac{\epsilon}{M}\right)) \cdot m(\left(a,a+\frac{\epsilon}{M}\right)) \leq M \cdot \frac{\epsilon}{M} = \epsilon$$
$$ \text{p.c} \int_{[a,b]} \overline{g} = \inf f(\left(a,a+\frac{\epsilon}{M}\right)) \cdot m(\left(a,a+\frac{\epsilon}{M}\right)) \geq -M \cdot \frac{\epsilon}{M} = -\epsilon$$
and similarly with $(b-\frac{\epsilon}{M},b)$.
\end{proof}

Now we have all the machinery to prove the theorem assertion. That is, let I be an interval. Let

f : I \to ℝ

be a bounded, piecewise continuous function. We demonstrate that

f

is Riemann integrable.

Let $I_1,\ldots,I_n$ be a partition of $I$ with respect to which $f$ is piecewise continuous. If on each of the intervals $I_i$ the function $f$ is Riemann integrable, then $f$ is Riemann integrable on the whole interval $I$ (e.g. taking the combination of piecewise upper and lower functions). Thus, it suffices to show that $f$ is Riemann integrable on some interval $I_i$, on which it is continuous and bounded. But no matter which boundaries $I_i$ includes, it can be partitioned into purely open and purely closed sets. By Lemma 1 and 2 of this proof, $f$ is Riemann integrable on each of these partitions. Thus, by the same argument $f$ Riemann integrable on the whole $I_i$, and we are done.

Exercise 1.1.23 (Bounded continuous functions are Riemann integrable)

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Exercise 1.1.23 (Bounded continuous functions are Riemann integrable)

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