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 $f,g:Atomathbb{R}$ 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

$$underline{int_A}{f(x) dx} = sup_{P} s(f,P) = sup_{P}
sum_{Bin P} m_bcdot vol B$$ $$overline{int_A}{f(x) dx} =
inf_{P} S(f,P) = sup_{P} sum_{Bin P} M_bcdot vol B$$

where $m_b = inf {f(x); xin B}$ 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 $mcdot vol A le s(f,P) le S(f,P)le Mcdot vol A$

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

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

where $m_B = inf {f(x), xin B}$, $m_B’ = inf {g(x), xin B}$, $m_B” = inf {f(x)+g(x), xin B}$. 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?

Answer

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).

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