Show that if $f,g: [a,b] \rightarrow \mathbb{R}$ are Riemann integrable then $\sqrt{f^2+g^2}$ is Riemann integrable and

$\int_a^b \sqrt{f^2 + g^2} dx \leq \int_a^b |f| dx + \int_a^b |g| dx$

So I'll start by writing down what I know:

I know by arithmetic of integrals that $|\int_a^b fg dx| \leq \sqrt{\int_a^b f^2} dx \cdot \sqrt{\int_a^b g^2 dx}$.

I also know that the function will be RI iff: $\exists P=\{a=x_0, x_1, ... , x_n = b\} \text{ a partition}$ s.t $U(f,p) - L(f,p) < \epsilon$. Finally, if f is any bounded function then $\overline{\int} f dx \geq \underline{\int} f dx$.

And so to prove Riemann integrability I'm guessing we let choose partitions such that:

$U(f,p_1) - (\text{ some } \epsilon ) < \int_a^b fdx < L(f,p_1) + (\text{ some } \epsilon)$. Similarly for $p_2$.

Let $P = P_1 \cup P_2$ band let $x,y \in [X_{i-1}, x_i]$ for some $1 \leq i \leq N$. And so we're left with

$\sqrt{(f^2(x) + g^2(x)) - ((f^2(y)-g^2(y))}$

I know I'm supposed to show that the above is less than something, and do an $\epsilon$ argument, but I really can't figure it out.

As for proving the inequality,

$\sqrt{U(f^2+g^2,p)} \leq \sqrt{\sum \sup f^2+g^2 \Delta x_i}$ And again, I don't know how to proceed.

Help would be much appreciated!


$f,g$ are Riemann-Integrable $\implies f^2,g^2$ are Riemann-Integrable

$f,g$ are Riemann-Integrable $\implies f+g $ are Riemann-Integrable

$ h$ is Riemann-Integrable $\text {and} h\ge 0 \implies \sqrt h$ is Riemann-Integrable

Now combine all three.

  • $\begingroup$ I think this is the simplest way to go. $\endgroup$ – zhw. Dec 8 '16 at 5:28

We have $\sqrt{a^2+b^2}\leq |a|+|b|$. See this simply by squaring each side. Now, monotonicity of the integral implies your result.

  • $\begingroup$ I'm probably being really stupid, but I'm still stuck. So we have $\sqrt{(f^2(x) + g^2(x)) - ((f^2(y)-g^2(y))} \leq |f(x)+g(x)| + |-f(y) + g(y)|$. I'm guessing I want to set a M = sup (something with f(x), g(x)) so we can further bound the above equation. $\endgroup$ – Nikitau Dec 8 '16 at 4:15
  • $\begingroup$ If you have Rudin's PMA, I'm suggesting you apply Theorem 6.12 (b) $\endgroup$ – Half-Pint Dec 8 '16 at 17:25
  • $\begingroup$ Yup, I do have Rudin's PMA. I was trying to mimic a proof I saw for 6.12(d), but I'll try figuring out (b). $\endgroup$ – Nikitau Dec 8 '16 at 18:59

Use Lebesgue criterion: $\sqrt{f^2+g^2}\le\lvert f\rvert+\lvert g\rvert$, which is bounded, and the discontinuities of $\sqrt{f^2+g^2}$ are a subset of the union of discontinuities of $f$ and $g$, which is a set of zero-measure.

BTW, it is superfluous to say that $f$ and $g$ are bounded. If they are Riemann integrable on a closed interval, they are automatically bounded.


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