If both $f$ and $g$ are not integrable, then $f+g$ and $fg$ are not integrable

Question

If both $f$ and $g$ are not integrable, then $f+g$ is not integrable

I think this is false. Take $f(x) = \begin{cases} 1 & \text{ if } x \in \mathbb{Q} \\ 0 & \text{ if } x \in \mathbb{Q}^c \end{cases}$ and $g(x) = \begin{cases} 1 & \text{ if } x \in \mathbb{Q}^c \\ 0 & \text{ if } x \in \mathbb{Q} \end{cases}$ Then $f+g = 1$, which is integrable.

If both $f$ and $g$ are not integrable, then $fg$ is not integrable. I think this is false, and I can take the same counterexample, and $fg = 0$, which is also integrable.

Are these correct?

If you mean integrable as in Riemann-integrable, then this seems correct! — Ulrik, Apr 08 '15 at 08:17
I think you are right. And the argument also applies to Lebesgue-integrablity, simply replace $\mathbb Q$ by any non mesurable set. — Censi LI, Apr 08 '15 at 08:21

score 0 · Answer 1 · answered Apr 08 '15 at 08:25

0

Both your counterexamples are correct.

answered Apr 08 '15 at 08:25

Jim

30,682

score 0 · Accepted Answer · answered Apr 08 '15 at 08:44

You examples are correct for Riemann integrability.

You can generalise your examples to Lebesgue integrability. Take, for example, any non-integable $f$ and put $g=-f$. Both $f$ and $g$ non-integrable, yet $f+g$ is integrable. You can also take two non-integrable functions $f$ and $g$ such their supports are disjoint (i.e. the sets where they are non-zero have empty intersection). Their product is identically zero, hence integrable.

If both $f$ and $g$ are not integrable, then $f+g$ and $fg$ are not integrable

2 Answers2