Show that $\overline{\int_a^b} f+g \le \overline{\int_a^b} f+\overline{\int_a^b}g$.

Question

Show that $\overline{\int_a^b} f+g \le \overline{\int_a^b} f+\overline{\int_a^b}g$.

Attempt: Using the fact that $\sup(x+y)\le \sup(x)+\sup(y),$

$$\inf (U(P,f+g))\le \inf(U(P,f)+U(P,g))$$ for any partition $P$ on $[a,b]$.

I am a little bit confused here. Is $\inf(U(P,f)+U(P,g))=\inf(U(P,f))+\inf(U(P,g))?$

Answer 1

No, but you can prove the result using common refinement. Let $\epsilon > 0$ and choose partitions $P$ and $Q$ such that $$U(P,f) < \overline{\int_a^b} f + \frac{\epsilon}{2}\quad \text{and}\quad U(Q,g) < \overline{\int_a^b} g + \frac{\epsilon}{2}$$

If $T$ be the common refinement of $P$ and $Q$, show that

$$U(T,f+g) \le \overline{\int_a^b} f + \overline{\int_a^b} g + \epsilon$$

using $U(T,f) \le U(P,f)$ and $U(T,g)\le U(Q,g)$. Then

$$\overline{\int_a^b} (f + g) \le \overline{\int_a^b} f + \overline{\int_a^b} g + \epsilon$$

Since $\epsilon$ was arbitrary, the result follows.

Answer 2

To prove by contradiction, assume that

$$\overline{\int_a^b} (f + g) > \overline{\int_a^b} f + \overline{\int_a^b} g.$$ Then there exists a partition $P$ such that $$\overline{\int_a^b} (f+g) - \overline{\int_a^b} g > U(P,f) \geqslant \overline{\int_a^b} f.$$

Since,

$$\overline{\int_a^b} (f+g) - U(P,f) > \overline{\int_a^b} g,$$ there exists a partition $P’$ such that $$\overline{\int_a^b} (f+g) - U(P,f) > U(P’,g) \geqslant \overline{\int_a^b} g,$$ and $$\overline{\int_a^b}(f+g) > U(P,f) + U(P’,g) .$$

Take a common refinement of the partitions $Q = P \cup P'$. Since upper sums decrease as partitions are refined, we have,

$$\tag{*}U(Q,f+g) \geqslant \overline{\int_a^b} (f+g) > U(P,f) + U(P’,g) \geqslant U(Q,f) + U(Q,g).$$

However, $\sup [f(x) + g(x)] \leqslant \sup f(x) + \sup g(x)$ and it follows that

$$U(Q,f+g) \leqslant U(Q,f) + U(Q,g),$$

which contradicts (*).