# Prove that $\underline{\int_A}{f(x) \ dx} + \underline{\int_A}{g(x) \ dx} \le \underline{\int_A}{[f(x) + g(x)] \ dx}$

Question:

Let $f,g:A\to\mathbb{R}$ bounded in the set $A$. Prove that

a)

$$\underline{\int_A}{f(x) \ dx} + \underline{\int_A}{g(x) \ dx} \le \underline{\int_A}{[f(x) + g(x)] \ dx}\\\le \overline{\int_A}{[f(x) + g(x)] \ dx} \le \overline{\int_A}{f(x) \ dx} + \overline{\int_A}{g(x) \ dx}$$

b) Give an example where all inequalities above are explicit

*

Remember that

$$\underline{\int_A}{f(x) \ dx} = \sup_{P} s(f,P) = \sup_{P} \sum_{B\in P} m_b\cdot vol B$$ $$\overline{\int_A}{f(x) \ dx} = \inf_{P} S(f,P) = \sup_{P} \sum_{B\in P} M_b\cdot vol B$$

where $m_b = \inf \{f(x); x\in B\}$ and $M_b = \sup \{f(x); x \in B\}$

The part $\underline{\int_A}{[f(x) + g(x)] \ dx}\le \overline{\int_A}{[f(x)+g(x)] \ dx}$ is obvious and comes from the fact that $m\cdot vol A \le s(f,P) \le S(f,P)\le M\cdot vol A$

Now for $\underline{\int_A}{f(x) \ dx} + \underline{\int_A}{g(x) \ dx} \le \underline{\int_A}{[f(x) + g(x)] \ dx}$, lets think:

It's

$$\sup s(f,P) + \sup s(g,P) \le \sup s(f+g, P) = \\ \sup \sum_{B\in P} m_B \cdot vol B + \sup \sum_{B\in P} m_B' \cdot vol B \le \sup \sum_{B\in P} m_B'' \cdot vol B$$

where $m_B = \inf \{f(x), x\in B\}$, $m_B' = \inf \{g(x), x\in B\}$, $m_B'' = \inf \{f(x)+g(x), x\in B\}$. How to proceed here?

For b), I can only find examples for the first or last inequality, not all at the same time. Somebody has an idea?

• What are your examples? Can't they be simply concatenated to get an example all the same time? – Berci Oct 7 '17 at 21:50

Assume that

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

and show this leads to a contradiction.

Making this assumption we have $$\underline{\int}_A [f(x)+g(x)] \, dx - \underline{\int}_A g(x) \,dx < \underline{\int}_A f(x) \, dx,$$ and there exists a partition $P$ and lower sum $L(P,f)$ such that $$\underline{\int}_A [f(x)+g(x)] \, dx - \underline{\int}_A g(x) \,dx < L(P,f) \leqslant \underline{\int}_A f(x) \, dx.$$

Hence,

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

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

$$\tag{*}L(Q,f+g) \leqslant \underline{\int}_A [f(x)+g(x)] \, dx < L(P,f) + L(P’,g) \leqslant L(Q,f) + L(Q,g).$$

However, $\inf f + \inf g \leqslant \inf (f+g)$ and, therefore, $L(Q,f) + L(Q,g)\leqslant L(Q,f+g)$ which is contradicted by (*).

A similar argument applies to the upper sum inequality. We have strict equality if $f$ and $g$ are Riemann integrable. We have strict inequality by choosing $A = [0,1]$ and $f = \chi_{\mathbb{Q} \cap [0,1]}$ and $g = -f$.

Problem with your reasoning for (a).

Since for any partition $P$ and lower sums we have $L(P,f) + L(P,g) \leqslant L(P,f+g)$ we can argue that

$$\tag{1}\sup_P (L(P,f) + L(P,g)) \leqslant \sup_P L(P,f+g) \leqslant \underline{\int}_A [f(x)+g(x)] \, dx$$

We also have that

$$\tag{2} \sup_P (L(P,f) + L(P,g)) \leqslant \sup_P L(P,f) + \sup_P L(P,g) \leqslant \underline{\int}_A f(x) \, dx + \underline{\int}_A g(x) \,dx$$

However from these results we cannot directly conclude anything about the relative ordering of the right-hand sides of (1) and (2).