# When stating the properties of integrals, why do we require their respective functions to be bounded in addition to being integrable?

When I learned about the properties of integrals, as the one below, for instance, I noticed the preamble of the theorem requires the respective functions to be both bounded and integrable.

If $$f$$ and $$g$$ are bounded, integrable functions on $$[a, b]$$, then so is $$f + g$$ and

$$\int_a^b (f(x) + g(x)) dx = \int_a^b f(x) dx + \int_a^b g(x) dx$$

My question is why do the functions have to be bounded? Isn't it enough for them to be integrable? I have not completed my Calculus course yet, so there are certain concepts I don't know about yet, like improper integrals.

• If the function is integrable on $[a,b]$ then it must also be bounded by definition. See also this question. – Peter Foreman Jul 30 '19 at 21:37

## 1 Answer

The previous comment is correct: integrable functions are bounded by definition. I think the reason that the word "bounded" is emphasized here does indeed have to do with improper integrals. For example, if $$\int^{b}_{a} f(x) dx$$ converges to $$+ \infty$$ and $$\int^{b}_{a} g(x) dx$$ converges to $$- \infty$$, then their sum is equal to $$\infty - \infty$$, which is an indeterminate form.

• What would $\int_a^b [f(x) + g(x)]\, dx$ evaluate to in the example you gave? Would it evaluate to $0$? – Calculemus Aug 24 '19 at 14:56