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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.

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

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

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