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Prove this:

Let $f :[a,b] \to \mathbb{R}$ be a bounded and integrable function. Show that $f^2$ is integrable too.

I'm in trouble with this. Can anyone show how to do it?

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  • $\begingroup$ What do you mean f is limited. Does it mean that $\exists M\in\mathbb{R}$ such that $|f(x)|\le M$ for all $x\in\mathbb{R}$? Is this with respect to the Riemann integral or the Lebesgue integral? $\endgroup$ – Laars Helenius Oct 13 '14 at 22:25
  • $\begingroup$ I changed “limited” to ”bounded”, which is the most commonly used term in English. $\endgroup$ – egreg Oct 13 '14 at 23:08
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Considering Riemann integrals, you can make the following argument.

Notice that if $f$ is bounded with $|f(x)| \leq B$, then

$$|f^2(x) - f^2(y)|= |f(x)-f(y)||f(x)+f(y)|\leq 2B|f(x)-f(y)|$$

and

$$M_j(f^2)-m_j(f^2)=\sup \{f^2(x):x_{j-1}\leq x\leq x_j\}-\inf \{f^2(x):x_{j-1}\leq x\leq x_j\}\leq 2B[M_j(f)-m_j(f)]$$

The difference of upper and lower sums for $f^2$ with respect to a partition $P$ is

$$U(P,f^2) - L(P,f^2) = \sum_{j=1}^{n}[M_j(f^2)-m_j(f^2)](x_j-x_{j-1}) \leq 2B[U(P,f) - L(P,f)].$$

Since $f$ is integrable, for any $\epsilon > 0$ there exists a partition P such that

$$U(P,f) - L(P,f) < \frac{\epsilon}{2B},$$

and

$$U(P,f^2) - L(P,f^2) < \epsilon.$$

Therefore, $f^2$ satisfies the Riemann criterion for integrability.

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If you are familiar with it, this follows immediately from Lesbegue Integrability Criteria.

As $f$ is bounded, so is $f^2$. Moreover, the set of discontinuities of $f^2$ is a subset of the set of discontinuities of $f$.

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  • $\begingroup$ how we can conclude that the set of discontinuities of $f^2$ is a subset the set of discontinities of $f$ ? $\endgroup$ – user2312512851 Oct 29 '17 at 16:49
  • $\begingroup$ @Photoneaterman This is equivalent to proving that if $f$ is continuous at $a$ then $f^2$ is continuous at $a$. $\endgroup$ – N. S. Oct 29 '17 at 18:12
  • $\begingroup$ but you've said that " subset " so I wonder how to find it. For example if we took a function (1 for rationals, -1 for irrationals) it is nowhere continuous but $f^2$ is everywhere continuous, as you've written the answer of yours, " the set of discontinuities of $f^2$ is a subset of the set of discontinuities of $f$" is being true. $\endgroup$ – user2312512851 Oct 29 '17 at 18:22
  • $\begingroup$ @Photoneaterman You don't "find" it. The statement "if $f$ is continuous at $a$ then $f^2$ is continuous at $a$" is equivalent to "if $f^2$ is discontinuous at $a$ then $f$ is discontinuous at $a$". This means exactly that the set of discontinuities is a subset. $\endgroup$ – N. S. Oct 29 '17 at 18:45

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