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- Apr 13, 2013

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I have to show that if $f:[a,b] \to \mathbb{R}$ is integrable,then $f^2$ is also integrable.

Knowing that $f:[a,b] \to \mathbb{R}$ is integrable,does this mean that $f$ is bounded??

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- Thread starter
- #1

- Apr 13, 2013

- 3,739

I have to show that if $f:[a,b] \to \mathbb{R}$ is integrable,then $f^2$ is also integrable.

Knowing that $f:[a,b] \to \mathbb{R}$ is integrable,does this mean that $f$ is bounded??

- Feb 13, 2012

- 1,704

If $f:[a,b]$ is Riemann integrable, then necessarly f(*) is bounded in [a,b]...Knowing that $f:[a,b] \to \mathbb{R}$ is integrable,does this mean that $f$ is bounded??

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$\chi$ $\sigma$

- Jan 26, 2012

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Hello.Knowing that $f:[a,b] \to \mathbb{R}$ is integrable,does this mean that $f$ is bounded??

There are two ways to define what it means to be "integrable" on $[a,b]$.

(i) The original Riemann definition: there exists a real number $I$ such that for any $\varepsilon > 0$, there is a $\delta > 0$ so that for any partition with norm $<\delta$ we have that all Riemann sums with respect to partition are within $\varepsilon$ of $I$.

(ii) The modified Darboux definition: the lower integral and the upper integral are equal.

In (i) there is no mention of boundness. In (ii) it is assumed that the function is bounded because we work with its sup and inf. But it is not hard to show that in (i) boundness is a necessary condition of an integrable function.

Which definition is it that you use?

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- Jan 26, 2012

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I have to show that if $f:[a,b] \to \mathbb{R}$ is integrable,then $f^2$ is also integrable.

Knowing that $f:[a,b] \to \mathbb{R}$ is integrable,does this mean that $f$ is bounded??

Alternatively, you could use the Riemann-Lebesgue Theorem, if you've gotten that far and are allowed to use it.Hello.

There are two ways to define what it means to be "integrable" on $[a,b]$.

(i) The original Riemann definition: there exists a real number $I$ such that for any $\varepsilon > 0$, there is a $\delta > 0$ so that for any partition with norm $<\delta$ we have that all Riemann sums with respect to partition are within $\varepsilon$ of $I$.

(ii) The modified Darboux definition: the lower integral and the upper integral are equal.

In (i) there is no mention of boundness. In (ii) it is assumed that the function is bounded because we work with its sup and inf. But it is not hard to show that in (i) boundness is a necessary condition of an integrable function.

Which definition is it that you use?

- Jan 26, 2012

- 236

I doubt you mean the one from Fourier analysis. So you probably mean the one that says a function is integrable if and only if its set of discontinuity is of measure zero. But this theorem says nothing about boundedness.Alternatively, you could use the Riemann-Lebesgue Theorem, if you've gotten that far and are allowed to use it.

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- Jan 26, 2012

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Right - I've usually heard that one called the "Riemann-LebesgueI doubt you mean the one from Fourier analysis.

Right.So you probably mean the one that says a function is integrable if and only if its set of discontinuity is of measure zero.

I agree, but it would fairly straight-forwardly prove that if $f$ is integrable, then $f^2$ is integrable. Boundedness requires some other machinery.But this theorem says nothing about boundedness.

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- Feb 7, 2012

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If $f$ is the function defined on the unit interval [0,1] by $f(x) = x^{-1/2}$ for $x>0$, and $f(0) = 0$, then \(\displaystyle \int_0^1f(x)\,dx\) exists as an improper Riemann integral, but $f^2$ is not integrable.

I have to show that if $f:[a,b] \to \mathbb{R}$ is integrable,then $f^2$ is also integrable.

Knowing that $f:[a,b] \to \mathbb{R}$ is integrable,does this mean that $f$ is bounded??

The definition of the Riemann integral requires that the function should be bounded. In order to deal with unbounded functions you need to use the improper Riemann integral. In that situation it is not true that the square of an integrable function is integrable, as the above example shows. Presumably your question refers to the ordinary Riemann integral for bounded functions.