Question

Prove that the function \(f(x)=\frac{1}{x}-\left\lfloor\frac{1}{x}\right\rfloor\) for \(x \neq 0\), \(f(0)=0\), is integrable on the closed interval \([0,1]\).

Short answer

Answer: Yes, because \(f(x)\) is both continuous and bounded on the interval \([0,1]\).

Step-by-step solution

Define the function

We are given the function \(f(x)=\frac{1}{x}-\left\lfloor\frac{1}{x}\right\rfloor\) for \(x \neq 0\), and \(f(0)=0\).

Identify the interval

The interval to prove integrability is \([0,1]\).

Show boundedness

We need to show that the function \(f(x)\) is bounded on the interval \([0,1]\). Consider the floor function portion \(\left\lfloor\frac{1}{x}\right\rfloor\). Since \(0 \leq x \leq 1\), we have \(\frac{1}{x} \geq 1\) for \(x > 0\). Therefore, \(\left\lfloor\frac{1}{x}\right\rfloor \geq 1\) for \(x > 0\). As a result, \(f(x) = \frac{1}{x} - \left\lfloor\frac{1}{x}\right\rfloor \leq \frac{1}{x} - 1 \leq 0\) for \(x > 0\). Since \(f(0) = 0\), we can conclude that \(f(x)\) is bounded on the interval \([0,1]\).

Show continuity

Now we want to prove that \(f(x)\) is continuous on the interval \([0,1]\). Since the function is defined piecewise, we must show continuity for each piece. For \(x \neq 0\), the function \(f(x) = \frac{1}{x}-\left\lfloor\frac{1}{x}\right\rfloor\) is the difference of two continuous functions, both of which are continuous on \((0, 1]\). Therefore, \(f(x)\) is continuous on \((0, 1]\). For \(x=0\), we can show continuity by the limit approach: \(\lim_{x \to 0^+}f(x) = \lim_{x \to 0^+}\left(\frac{1}{x}-\left\lfloor\frac{1}{x}\right\rfloor\right) = \lim_{x \to 0^+}\left(\frac{1}{x} - 1\right) = -1 + 1 = 0\). Therefore, the one-sided limit at \(x=0\) equals \(f(0) = 0\), and the function is continuous at \(x=0\).

Conclude integrability

Since we have shown that \(f(x)\) is both continuous and bounded on the interval \([0,1]\), we can conclude that \(f(x)\) is integrable on the closed interval \([0,1]\).