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

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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?

Assume that

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

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).