From "An Introduction to Lebesgue Integration and Fourier Series" by Howard J. Wilcox and David L. Myers:
1.1 Definition: A partition $P$ of a closed interval $[a, b]$ is a finite sequence $(x_{0}, x_{1}, \ldots, x_{n})$ such that $a = x_{0} < x_{1} < \ldots < x_{n} = b$. The norm of $P$, denoted $\left|\left|P\right|\right|$, is defined by $\left|\left|P\right|\right| = \max_{1 \leq i \leq n} (x_{i} – x_{i-1})$.
1.2 Definition: Let $P = (x_{0}, \ldots, x_{n})$ be a partition of $[a, b]$, and let $f$ be defined on $[a, b]$. For each $i = 1, \ldots, n$, let $x_{i}*$ be an arbitrary point in the interval $[x_{i-1}, x_{i}]$. Then any sum of the form $R(f, P) = \sum_{i=1}^{n} f(x_{i}*)(x_{i} – x_{i-1})$ is called a Riemann sum of $f$ relative to $P$.
1.3 Definition: A function $f$ is Riemann integrable on $[a, b]$ if there is a real number $R$ such that for any $\epsilon > 0$, there exists a $\delta > 0$ such that for any partition $P$ of $[a, b]$ satisfying $\left|\left|P\right|\right| < \delta$, and for any Riemann sum $R(f, P)$ of $f$ relative to $P$, we have $\left|R(f,P) – R\right| < \epsilon$.
1.4 Definition: A function $g$, defined on $[a, b]$, is a step function if there is a partition $P = (x_{0}, x_{1}, \ldots, x_{n})$ such that $g$ is constant on each open subinterval $(x_{i-1}, x_{i})$, for $i = 1, \ldots, n$.
1.5 Proposition: Any step function $g$ on $[a, b]$ is Riemann integrable. Furthermore, if $g(x) = c_{i}$ for $x \in (x_{i-1}, x_{i})$, where $(x_{0}, \ldots, x_{n})$ is a partition of $[a, b]$, then $\int_{a}^{b} g(x) dx = \sum_{i=1}^{n} c_{i}(x_{i} – x_{i-1})$.
I'm trying to prove proposition 1.5. I can probably do so, but it seems like I need to go through a lot of different cases for the points $x_{0}, \ldots, x_{n}$. Is there a simpler approach? The proofs I can find use a different definition of the Riemann integral. I can prove they are equivalent, but I need this proposition to do so.
Best Answer
Scratch work / thinking about the problem...
There are $n-1$ points where the value of the step function can step. For a generic partition (into many segments), $P = (p_0 = a, \dots, p_{\text{big number}} = b)$, the partition elements contained on the interiors of the steps contribute a known quantity to the Riemann sum and the partition elements that contain the $x_i$ ($0 < i < n$) may contribute different amounts to the sum, depending on whether the sample point (the $x_i^*$s) is selected from the left or the right of the $x_i$. This means we have $n-1$ summands in the Riemann sum that we don't know their contribution. But we can bound them. The largest magnitude from each would be $\pm ||P|| \max_i |c_i|$. Since there are $n-1$ of them, we need $(n-1)||P|| \max_i |c_i| < \varepsilon$, so equivalently, $||P|| < \frac{\varepsilon}{(n-1) \max_i |c_i|}$, which means we know to what to pick for $\delta$ to apply Definition 1.3.
Writing up ...
Applying Definition 1.3, let $R = \sum_{i=1}^n c_i(x_i - x_{i-1})$ and $\varepsilon > 0$ be given. Set $\delta = \frac{\varepsilon}{(n-1) \max_i |c_i|}$ and let, for $m > 0$, $P = (a = p_0, \dots, p_m = b)$ be any partition satisfying $||P|| < \delta$. Then for $R(g,P)$, a Riemann sum of $g$ relative to $P$, we have \begin{align*} |R(g,P) - R| &< \dots \\ &< \text{[insert sneaky use of triangle inequality here]} \\ &< \dots \\ &< \varepsilon \text{.} \end{align*} This shows that $g$ is Riemann integrable.
Continuing ...
You haven't quoted anything defining the symbol $\int$ or assigning values to expressions containing such a symbol. Guessing about what those missing parts would contain, we have shown that $$ R - (n-1)||P|| \max_i |c_i| < R(g,P) < R + (n-1)||P|| \max_i |c_i| \text{.} $$ We have $n$ and the $c_i$ fixed by the statement of the proposition. In the limit as $||P|| \rightarrow 0$, both the lower and upper bound in this inequality tend to $R$, so $R(g,P) \rightarrow R$.
Let $||x|| = \min \{x_{i+1} - x_i : i = 0 \dots n-1\}$. We are interestd in the limit as $||P|| \rightarrow 0$, so we may require $||P|| < (1/2)||x||$. This means no element of the partition $P$ contains two of the $x_i$. In particular, we may collect the intervals in $P$ into two sets, $I$, the set of partition members which do not contain an $x_i$ (for $i = 1 \dots n-1$) and $J$, the set of partition members that do contain an $x_i$ (for $i = 1 \dots n-1$). So $|J| = m - (n-1)$ and $|I| = n-1$. Note that $P = I \cup J$.
(Definition 1.4 should contain the constraint $x_0 = a$ and $x_n = b$, otherwise $g$ need not be constant on the first and last pieces of the interval $[a,b]$. This means we don't care about "jumps" at $x_0$ and $x_n$ since the value(s) left of $x_0$ and right of $x_n$ are invisible to an integral on $[a,b]$.)
The contributions to $R(g,P)$ from members of $I$ are easy to write down: (width of partition element)(height of function, constant on that interval). Since the height is constant on that interval, it does not matter where the sample point is chosen.
The contributions to $R(g,P)$ from members of $J$ are each one of two values. For $i=1 \dots n-1$, let $q_i = [q_{i,0},q_{i,1}] \in J$ be the element of $P$ containing $x_i$, so $q_{i,0} \leq x_i < q_{i,1}$. The contributions to $R(g,P)$ from each of these partition elements is either $$ (q_{i,1} - q_{1,0})c_i \text{,} $$ $$ (q_{i,1} - q_{1,0})g(x_i) \text{, or} $$ $$ (q_{i,1} - q_{1,0})c_{i+1} \text{,} $$ depending on whether the sample point from this partition element is taken from the left of $x_i$, from $x_i$, or from the right of $x_i$.
Let $M = \max \left( \{|c_i|\}_i \cup \{|g(x_i)|\}_i \right)$. Then the contribution from this element of $J$ is bounded above by $M (q_{i,1} - q_{1,0})$ and below by $-M(q_{i,1} - q_{1,0})$. So the total contribution by $J$ to any $R(g,P)$ is in the interval $$ \left[ -(n-1)M(q_{i,1} - q_{1,0}), (n-1)M(q_{i,1} - q_{1,0}) \right] $$ which $\rightarrow 0$ as $||P|| \rightarrow 0$.