Proof that if $f$ is integrable then also $f^2$ is integrable

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.


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 $f$.

Tags:

Integration

Real Analysis

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