Problem 6 - A Course in Calculus and Real Analysis

Question

Let \(f, g:[a, b] \rightarrow \mathbb{R}\) be bounded functions. Show that $$ L(f)+L(g) \leq L(f+g) \quad \text { and } \quad U(f+g) \leq U(f)+U(g) $$ Hence conclude that if \(f\) and \(g\) are integrable, then so is \(f+g\), and the Riemann integral of \(f+g\) is equal to the sum of the Riemann integrals of \(f\) and \(g\).

Step-by-Step Solution

Answer

We can prove the desired result following these steps: 1. Define lower and upper sums of the functions \(f, g\) and \(f+g\). 2. Prove that \(L_P(f) + L_P(g) \leq L_P(f+g)\) and \(U_P(f+g) \leq U_P(f) + U_P(g)\). 3. Show that if \(f\) and \(g\) are integrable, then their sum \(f+g\) is also integrable. 4. Finally, prove that \(\int_a^b (f+g)(x)dx = \int_a^b f(x)dx + \int_a^b g(x)dx\), which means that the Riemann integral of \(f+g\) equals the sum of the integrals of \(f\) and \(g\).

Step 1: Define lower sums and upper sums of functions

Let \(f, g : [a, b] \rightarrow \mathbb{R}\) be bounded functions, and let \(P = \{x_0, x_1, ..., x_n \}\) be a partition of the interval \([a, b]\). For simplicity, we denote the iterated minimum of \(f\) and \(g\) on the interval \([x_{i-1}, x_i]\) by \(m_i(f)\) and \(m_i(g)\), respectively, and the iterated maximum by \(M_i(f)\) and \(M_i(g)\), respectively. Then, the lower sums of \(f, g\) and \(f+g\) on the partition \(P\) are defined as follows: \[L_P(f) = \sum_{i=1}^n m_i(f)(x_i - x_{i-1}),\] \[L_P(g) = \sum_{i=1}^n m_i(g)(x_i - x_{i-1}),\] \[L_P(f+g) = \sum_{i=1}^n m_i(f+g)(x_i - x_{i-1}).\] Similarly, the upper sums of \(f, g\) and \(f+g\) on the partition \(P\) are defined as: \[U_P(f) = \sum_{i=1}^n M_i(f)(x_i - x_{i-1}),\] \[U_P(g) = \sum_{i=1}^n M_i(g)(x_i - x_{i-1}),\] \[U_P(f+g) = \sum_{i=1}^n M_i(f+g)(x_i - x_{i-1}).\]

Step 2: Prove the required inequalities

By definition, we have: \( m_i(f) + m_i(g) \leq m_i(f+g) \leq M_i(f) + M_i(g) \) for each \(i = 1, 2, ..., n\) Now we will derive the inequalities using these relationships: \[L_P(f) + L_P(g) = \sum_{i=1}^n (m_i(f) + m_i(g))(x_i - x_{i-1}) \leq \sum_{i=1}^n m_i(f+g)(x_i - x_{i-1}) = L_P(f+g)\] \[U_P(f) + U_P(g) = \sum_{i=1}^n (M_i(f) + M_i(g))(x_i - x_{i-1}) \geq \sum_{i=1}^n M_i(f+g)(x_i - x_{i-1}) = U_P(f+g)\] Thus, we have established the required inequalities.

Step 3: Prove the integrability of the sum of functions

Since \(f\) and \(g\) are integrable, we know that for any given \(\epsilon > 0\), there exists a partition \(P\) of \([a, b]\) such that: \[U_P(f) - L_P(f) < \frac{\epsilon}{2}\] and \[U_P(g) - L_P(g) < \frac{\epsilon}{2}\] Using the inequalities established in step 2, we have: \[U_P(f+g) - L_P(f+g) \leq (U_P(f) + U_P(g)) - (L_P(f) + L_P(g)) \leq U_P(f) - L_P(f) + U_P(g) - L_P(g) < \epsilon\] Since we have found a partition \(P\) of \([a, b]\) such that the difference between upper and lower sums of \(f+g\) is smaller than an arbitrary positive \(\epsilon\), we conclude that the sum of the functions, \(f+g\), is also integrable.

Step 4: Prove the equality of the sum of integrals

From step 3, we know that both the lower and upper integral of the sum of the functions exists. We need to show that their sum equals that of the individual integrals: \[\int_a^b (f+g)(x)dx = \int_a^b f(x)dx + \int_a^b g(x)dx\] Since \(f\) and \(g\) are integrable, the lower integrals of \(f\) and \(g\) equal their upper integrals. So from the inequalities established in step 2, we have: \[\int_a^b (f+g)(x)dx \geq \int_a^b f(x)dx + \int_a^b g(x)dx\] On the other hand, if we change \(g\) to \(-g\) and repeat the same argument, we get: \[\int_a^b (f-g)(x)dx \geq \int_a^b f(x)dx - \int_a^b g(x)dx\] Thus, we have: \[\int_a^b g(x)dx \geq \int_a^b (f+g)(x)dx - \int_a^b f(x)dx\] This implies: \[\int_a^b (f+g)(x)dx \leq \int_a^b f(x)dx + \int_a^b g(x)dx\] Finally, we have the required equality: \[\int_a^b (f+g)(x)dx = \int_a^b f(x)dx + \int_a^b g(x)dx\] As a result, we have proven that if two functions are integrable, their sum is also integrable, and the Riemann integral of their sum equals the sum of their individual Riemann integrals.