I am working through Measure, Integration and Real Analysis by Sheldon Axler. I'm quite rusty with analysis and I've gotten stuck on this problem and would appreciate any help. Sorry if I made any obvious mistakes, analysis really isn't my strong suit.
Suppose $a \leq s < t \leq b$. Define $f: [a,b] \rightarrow \mathcal{R}$ by
$$ f(x)=
\begin{cases}
1& \text{if } s < x < t,\\
0 & \text{otherwise}.
\end{cases}$$
Prove that $f$ is Riemann integrable on $[a,b]$ and that $\int_a^bf = t -s$.
Here is the definition of Riemann integrable functions provided:
i) A bounded function on a closed bounded interval is called Riemann integrable if its lower Riemann integral equals its upper Riemann integral.
ii) If $f: [a,b] \rightarrow \mathcal{R}$ is Riemann integrable, then the Riemann integral $\int_a^bf$ is defined by
$\int_a^bf = L(f, [a,b]) = U(f, [a,b])$.
Finally, upper and lower Riemann integrals are defined as follows:
Suppose $f: [a,b] \rightarrow \mathcal{R}$ is a bounded function. The lower Riemann integral $L(f, [a,b])$ and the upper Riemann integral $U(f, [a,b])$ of $f$ are defined by
$$L(f, [a,b]) = \sup_{P} L(f, P, [a,b])$$, where P is a partition of $[a,b]$.
and
$$U(f, [a,b]) = \inf_{P} U(f, P, [a,b])$$
where the supremum and infimum are taken over all partitions $P$ of $[a,b]$.
Here's my attempt so far (again, sorry if I'm way off). If so, any guidance would be really appreciated as I'm very confused on how to solve this:
Let $P$ be the equally spaced partition $a=x_0, \dots, x_n=b$ of $[a,b]$ with
$$x_j – x_{j-1} = \frac{b-a}{n}$$
for each $j=1, \dots, n$. Then
$$\begin{align}
U(f, [a,b]) – L(f, [a,b]) &\leq U(f, P, [a,b]) – L(f, P, [a,b]) \\
&=\frac{b-a}{n} \sum_{j=1}^n\left( \sup_{[x_{j-1} – x_j]} f – \inf_{[x_{j-1} – x_j]} \right) \\
&\leq \frac{b-a}{n} \cdot t-s
\end{align}$$
where the RHS of the final inequality will approach 0 as $n$ approaches infinity (this is the part I'm really not sure about). This implies that $U(f, [a,b]) \leq L(f, [a,b])$.
By an earlier theorem in the book, we know that $L(f, [a,b]) \leq U(f, [a,b])$.
Therefore, we have that $f$ is Riemann integrable since $L(f, [a,b]) = U(f, [a,b])$..
That's what I have so far. Thanks for the help! Let me know if you need any more information or definitions.
Best Answer
Your estimate for $U(f, P, [a,b]) - L(f, P, [a,b])$ is not correct. It could be \begin{align} U(f, P, [a,b]) - L(f, P, [a,b]) &= \frac{b - a}{n}\sum_{j = 1}^{n}(\sup_{[x_{j - 1}, x_j]}f - \inf_{[x_{j - 1}, x_j]}f) \\ &\leq \frac{b - a}{n}\sum_{j = 1}^{n}(1 - 0) \\ &= b - a. \end{align} This estimate is too crude.
For proving Riemann integrability of $f$, it is useful to use the following result: A bounded function $g \colon [a, b] \to \mathbb{R}$ is Riemann integrable if and only if for every $\varepsilon > 0$, there exists a partition $P$ of $[a, b]$ such that $U(g, P) - L(g, P) \leq \varepsilon$.
Given $\varepsilon$, the general strategy to find $P$ such that $U(f, P) - L(f, P) \leq \varepsilon$ is to choose $P$ so that the portions of $[a, b]$ where $f$ is badly behaved (i.e. discontinuous) are squeezed in intervals of total length arbitrary small, e.g. $\leq K\varepsilon^p$ for some $K, p > 0$.