Problem 9 - A Course in Calculus and Real Analysis

Question

Let \(f:[a, b] \rightarrow \mathbb{R}\) be a function. Show that \(f\) is integrable if (i) \(r f\) is integrable for some nonzero \(r \in \mathbb{R}\), or (ii) if \(f\) is bounded, \(f(x) \neq 0\) for all \(x \in[a, b]\), and \(1 / f\) is integrable.

Step-by-Step Solution

Answer

In summary, we have shown that the function \(f(x)\) is integrable if either (i) \(rf(x)\) is integrable for a nonzero constant \(r\) or (ii) \(f(x)\) is bounded, nonzero on \([a, b]\), and \(1/f\) is integrable. For condition (i), we showed that factoring out the constant \(r\) from the integral proves integrability. For condition (ii), we defined function \(g(x) = \frac{1}{f(x)}\), then showed that the integral of \(g(x)f(x)\) exists, guaranteeing the integrability of \(f(x)\).

Step 1: Prove that if r*f is integrable so is f for nonzero r.

If \(rf\) is integrable, it means that \(\int_{a}^{b}rf(x) \, dx\) exists. Since \(r\) is just a nonzero constant, we can factor it out of the integral. \[\int_{a}^{b}rf(x) \, dx = r \int_{a}^{b}f(x) \, dx\] Now we need to prove that \(f\) is integrable. Since we can factor out the constant \(r\), we can say that the integral of \(f\) exists. Hence, we can conclude that if \(rf(x)\) is integrable, the function \(f(x)\) is also integrable. #Condition (ii)#

Step 2: Show that if f is bounded and nonzero, and 1/f is integrable, then f is integrable.

We are given that \(f\) is bounded, which means there exist constants \(m,M \in \mathbb{R}\) such that: \[m \leq f(x) \leq M, \quad \forall x \in [a,b]\] We are also given that \(f(x) \neq 0\) for all \(x \in [a,b]\), and that \(1/f\) is integrable. It means that \(\int_{a}^{b} \frac{1}{f(x)} \, dx\) exists. To prove that f is integrable, define another function \(g(x) = \frac{1}{f(x)}\) for \(x \in [a,b]\). Now we wish to find if the integral of \(g(x)\) multiplied by f(x) exists. Consider the integral \(\int_{a}^{b} g(x)f(x) \, dx\). Since \(g(x) = \frac{1}{f(x)}\), we have: \[\int_{a}^{b} g(x)f(x) \, dx = \int_{a}^{b} (1) \, dx\]

Step 3: Show f is integrable by showing the integral of g(x)f(x) exists.

Since the integral \(\int_{a}^{b} (1) \, dx\) exists and simply equals \(b-a\), we can conclude that the integrability of \(g(x)f(x)\) is guaranteed. So, if \(1/f\) is integrable, the function \(f(x)\) is also integrable. In conclusion, under two different conditions (i) and (ii), we have proven that the function \(f(x)\) is integrable on the given interval \([a, b]\).