if f is a riemann integrable on [a,b], then |f| is a riemann integrable on [a,b]. proof real analysis

### 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  
\[
|f(x)| =
\begin{cases}
f(x) & \text{if } f(x) \geq 0, \\
-f(x) & \text{if } f(x) < 0.
\end{cases}
\]  
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.

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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.

if f is riemann integrable on [a,b], then |f| is riemann integrable on [a,b]. proof real analysis

Explore the detailed proof that if f is Riemann integrable on [a,b], then its absolute value |f| is also Riemann integrable. This proof includes key concepts from real analysis, such as the Riemann Integrability Criterion and the preservation of integrability under the absolute value function.

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