Theorem:

If $f$ is Riemann integrable on the interval $[a, b]$, then $|f|$ is also Riemann integrable on $[a, b]$.

Proof:

1. Riemann Integrability Criterion
A function $f$ is Riemann integrable on $[a, b]$ if for any given $\epsilon > 0$, there exists a partition $P$ of $[a, b]$ such that the difference between the upper sum $U(f, P)$ and the lower sum $L(f, P)$ is less than $\epsilon$. Formally,
$\text{For every } \epsilon > 0, \text{ there exists a partition } P \text{ of } [a, b] \text{ such that } U(f, P) - L(f, P) < \epsilon.$

2. Properties of Absolute Value
Let $f$ be Riemann integrable on $[a, b]$. The function $|f|$ is defined as

\begin{cases} f(x) & \text{if } f(x) \geq 0, \\ -f(x) & \text{if } f(x) < 0. \end{cases}$$ Since the absolute value function is continuous, and composition of a Riemann integrable function $$f$$ with a continuous function (in this case, $$|\cdot|$$) preserves Riemann integrability, it is intuitive that $$|f|$$ should also be Riemann integrable. However, we'll give a more detailed argument below. **3. Upper and Lower Sums for $$|f|$$** Let $$P$$ be a partition of $$[a, b]$$, and let $$M_i$$ and $$m_i$$ denote the supremum and infimum of $$f(x)$$ over the $$i$$-th subinterval $$[x_{i-1}, x_i]$$. Then the upper sum and lower sum for $$f$$ are given by: $$U(f, P) = \sum_{i=1}^{n} M_i (x_i - x_{i-1}), \quad L(f, P) = \sum_{i=1}^{n} m_i (x_i - x_{i-1}).$$ For the function $$|f|$$, let $$M_i^*$$ and $$m_i^*$$ denote the supremum and infimum of $$|f(x)|$$ over $$[x_{i-1}, x_i]$$. Since $$|f(x)| \leq |M_i|$$ and $$|f(x)| \geq |m_i|$$, the upper and lower sums for $$|f|$$ can be written as: $$U(|f|, P) = \sum_{i=1}^{n} M_i^* (x_i - x_{i-1}), \quad L(|f|, P) = \sum_{i=1}^{n} m_i^* (x_i - x_{i-1}).$$ **4. Estimating the Difference in Upper and Lower Sums** We now show that the difference $$U(|f|, P) - L(|f|, P)$$ can be made arbitrarily small by choosing a fine enough partition $$P$$. Note that for any partition $$P$$, $$|f(x)| \leq |M_i| \quad \text{and} \quad |f(x)| \geq |m_i|$$ imply that $$U(|f|, P) \leq U(f, P)$$ and $$L(|f|, P) \geq L(f, P)$$. Thus, $$U(|f|, P) - L(|f|, P) \leq U(f, P) - L(f, P).$$ Since $$f$$ is Riemann integrable, for any $$\epsilon > 0$$, there exists a partition $$P$$ such that $$U(f, P) - L(f, P) < \epsilon.$$ Therefore, for this same partition $$P$$, we have: $$U(|f|, P) - L(|f|, P) \leq U(f, P) - L(f, P) < \epsilon.$$ **5. Conclusion** Since for any $$\epsilon > 0$$, we can find a partition $$P$$ such that $$U(|f|, P) - L(|f|, P) < \epsilon$$, this implies that $$|f|$$ is Riemann integrable on $$[a, b]$$. Thus, the theorem is proven. --- Would you like more details on any step or further clarifications? ### Related Questions: 1. What are the conditions for a function to be Riemann integrable? 2. How does the composition of functions affect Riemann integrability? 3. Why does the absolute value function preserve integrability? 4. Can a non-integrable function have its absolute value be integrable? 5. What is the relationship between the upper and lower sums in Riemann integration? ### Tip: A bounded function on a closed interval is Riemann integrable if the set of its discontinuities has measure zero.