Question

Can one assert that if a function is absolutely integrable on the interval \([\mathrm{a}, \mathrm{b}]\), then it is integrable on this interval ?

Short answer

Answer: Yes, if a function is absolutely integrable on an interval, it is also integrable on that interval.

Step-by-step solution

Definition of absolute integrability

First, let's define absolute integrability. A function \(f(x)\) is considered absolutely integrable on an interval \([\mathrm{a}, \mathrm{b}]\) if the integral of its absolute value exists and is finite: $$\int_a^b |f(x)| dx < \infty$$

Definition of integrability

Now, let's recall the definition of integrability. A function \(f(x)\) is considered integrable on an interval \([\mathrm{a}, \mathrm{b}]\) if the Riemann integral exists and is finite. This means that the difference between the upper and lower sums converge to zero: $$\lim_{||P|| \to 0} (U(P,f) - L(P,f)) = 0$$ where \(||P||\) is the maximum length of the subintervals in the partition, and \(U(P,f)\) and \(L(P,f)\) are the upper and lower sums, respectively.

Comparing absolute integrability and integrability

To prove the statement, we will use the following fact: a function \(f(x)\) is Riemann integrable on an interval \([\mathrm{a}, \mathrm{b}]\) if and only if it is bounded and its set of discontinuities has a measure of zero. Now, if \(f(x)\) is absolutely integrable, that means \(\int_a^b |f(x)| dx < \infty\). As the integral of the absolute value of \(f(x)\) is finite, it implies that \(f(x)\) is bounded on the interval \([\mathrm{a}, \mathrm{b}]\). Moreover, because \(|f(x)|\) is integrable, any point of discontinuity in \(f(x)\) must also have a measure of zero (otherwise the integral of the absolute value would not be finite).

Conclusion

We have demonstrated that if a function \(f(x)\) is absolutely integrable on an interval \([\mathrm{a}, \mathrm{b}]\), it must be bounded, and its set of discontinuities has a measure of zero. This means that \(f(x)\) is also Riemann integrable on this interval. As a result, the initial statement is true. If a function is absolutely integrable on an interval, it is also integrable on that interval.