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I'm trying to follow my textbook's prove for the following:
Let $f,g: [a,b] \rightarrow \mathbb{R} $ be integrable. Then $f+g$ is integrable and $ \int^{b}_{a}(f+g) = \int^{b}_{a}f + \int^{b}_{a}g $.

The book basically shows that for any partition $P$, the lower sum $L(f+g,[a,b],P) = L(f,[a,b],P) + L(g,[a,b],P) $. This I understand.

But from this it concludes, $ \underline\int^{b}_{a}(f+g) \leq \underline\int^{b}_{a}f + \underline\int^{b}_{a}g$ without any explanation.

Why is this the case? It seems to have skipped some steps.

user308485
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    what does $\underline\int$ mean ? and you say $f,g$ integrable but don't you consider at first $f,g \ge 0$, and even increasing ? (so that you can easily bound $L(f,[c,d],P)$ in term of $(d-c) f(d)$ and $(d-c) f(c)$) – reuns Mar 22 '16 at 01:05

2 Answers2

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We have $\inf_{[a,b]} (f+g)\geq \inf_{[a,b]} f+\inf_{[a,b]} g$.

Now, let $\mathcal P$ be the set of partitions of $[a,b]$. Then, we have, for $P\in \mathcal P$,

$L(f+g,P) \geq L(f,P) + L(g,P)$,

There are partitions $Q$ and $R$ and their common refinement $P$ s.t.

$\underline{\int}_a^bf-\frac{\epsilon }{2} <L(f,Q)\leq L(f,P)$ and $\underline{\int}_a^bf-\frac{\epsilon }{2} < L(f,R)\leq L(f,P)$.

Then, $\underline{\int}_a^b(f + g)\geq L(f+g,P)\geq L(f,P) + L(g,P)\geq \underline{\int}_a^bf-\frac{\epsilon }{2}+\underline{\int}_a^bg-\frac{\epsilon }{2}=\underline{\int}_a^bf+\underline{\int}_a^bg -\epsilon$

from which it follows that $\underline{\int}_a^b(f + g)\geq \underline{\int}_a^bf+\underline{\int}_a^bg $, which is the desired inequality.

Matematleta
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It's going to be impossible to prove that $f + g$ is integrable this way.

You ultimately want to show that

$$\underline{\int}_a^bf + \underline{\int}_a^bg \leqslant \underline{\int}_a^b(f + g) \leqslant \overline{\int}_a^b(f + g) \leqslant \overline{\int}_a^bf + \overline{\int}_a^bg ,$$

in order to show that integrability of $f + g$ follows from that of $f$ and $g$.

Note that on any subinterval $I$ of a partition

$$\inf_{x \in I}f(x) + \inf_{x \in I}g(x) \leqslant \inf_{x \in I}[f(x) + g(x)],$$

since for all $x \in I$,

$$f(x) + g(x) \geqslant \inf_{x \in I}f(x) + \inf_{x \in I}g(x).$$

Hence, forming lower sums we get,

$$L(f,P) + L(g,P) \leqslant L(f+g,P).$$

This is the correct starting point.

Assume that

$$\underline{\int}_a^b [f(x)+g(x)] \, dx < \underline{\int}_a^b f(x) \, dx + \underline{\int}_a^b g(x) \,dx ,$$

and show that this leads to a contradiction as follows.

Assuming this were true we have

$$\underline{\int}_a^b [f(x)+g(x)] \, dx - \underline{\int}_a^b g(x) \,dx < \underline{\int}_a^b f(x) \, dx,$$ and there exists a partition $P$ such that $$\underline{\int}_a^b [f(x)+g(x)] \, dx - \underline{\int}_a^b g(x) \,dx < L(P,f) \leqslant \underline{\int}_a^b f(x) \, dx.$$

Hence,

$$\underline{\int}_a^b [f(x)+g(x)] \, dx - L(P,f) < \underline{\int}_a^b g(x) \, dx,$$ and there exists a partition $P’$ such that $$\underline{\int}_a^b [f(x)+g(x)] \, dx - L(P,f) < L(P’,g) \leqslant \underline{\int}_a^b g(x) \, dx,$$ and $$\underline{\int}_a^b [f(x)+g(x)] \, dx < L(P,f) + L(P’,g) .$$

Now take a common refinement of the partitions $Q = P \cup P'$. Lower sums increase as partitions are refined and we have $L(Q,f) \geqslant L(P,f)$ and $L(Q,g) \geqslant L(P’,g).$

It follows that

$$L(Q,f+g) \leqslant \underline{\int}_a^b [f(x)+g(x)] \, dx < L(P,f) + L(P’,g) \leqslant L(Q,f) + L(Q,g).$$

This contradicts the inequality for lower sums, and, therefore

$$\underline{\int}_a^b f(x) \, dx + \underline{\int}_a^b g(x) \, dx \leqslant \underline{\int}_a^b [f(x) + g(x)] \, dx. $$

RRL
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  • Great answer. Seems strange that my text left out such important details. – user308485 Mar 22 '16 at 03:08
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    @user308485: You're very welcome. Its an interesting exercise to find non-Riemann integrable functions where the strict inequality holds. – RRL Mar 22 '16 at 03:12
  • @RRL : what $\underline{\int}$ and $\overline{\int}$ means – reuns Mar 22 '16 at 04:25
  • @user1952009: They are the lower and upper Darboux integrals meaning: $\underline{\int}a^b f := \sup{P} L(f,P)$ and $\overline{\int}a^b f := \inf{P} U(f,P)$. The lower integral is the least upper bound of lower sums formed from every possible partition. In the proof I use that if $A < \underline{\int}_a^b f$ then there must be a partition $P$ such that $A < L(f,P)$ since then $A$ cannot be an upper bound. – RRL Mar 22 '16 at 04:39
  • For a bounded function we always have $\underline{\int} f \leqslant \overline{\int} f$. A function is integrable iff $\underline{\int} f = \overline{\int} f$. – RRL Mar 22 '16 at 04:43
  • No. Pick any partition and form a lower sum. As you add points and refine the partition, the lower sum must increase -- basically because $\inf_{[x,y]} f \leqslant \inf_{[x,z]} f + \inf_{[z,y]} f$ if $x < z < y$. You can also show that the lower sums must be bounded above by any upper sum. Hence the supremum exists and is finite. – RRL Mar 22 '16 at 04:58