An assignment in Riemann's integral: Given two bounded function f and g in [a,b], if I proved that for every partition p: $$ U(f+g,p) \leq U(f,p)+U(g,p) $$
How can I show that:
I don't know how to format this correctly but my intention is upper Riemann integral $$ \int_a^b f+g \leq \int_a^b f + \int_a^b g $$
I understand intuitively that if I proved the inequality for every p then it should also be true for the infimums of the groups holding the upper sums, but for some reason I can't manage to prove it on paper, something doesn't add up.
Would appreciate help on how to prove it formally.
Take any partitions $P_1$ and $P_2$ of $[a,b]$. Let $P = P_1 \cup P_2$ be the common refinement. We then have (using your first result)
$$U(f+g,P) \leqslant U(f,P) + U(g,P) \leqslant U(f,P_1) + U(g,P_2)$$
The upper (Darboux) integral of $f+g$ is given by
$$\overline{\int_a^b} (f+g) = \inf_{P' \in \mathcal{P}}\,U(f+g,P'),$$
wher $\mathcal {P}$ is the class of all partitions of $[a,b]$.
Thus,
$$\overline{\int_a^b} (f+g) \leqslant U(f+g,P) \leqslant U(f,P_1) + U(g,P_2)$$
Taking the infimum on the RHS over all $P_1 \in \mathcal{P}$ followed by the infimum over all $P_2 \in \mathcal{P}$, we get
$$\overline{\int_a^b} (f+g) \leqslant U(f+g,P) \leqslant \inf_{P_1 \in \mathcal{P}}U(f,P_1) + \inf_{P_2 \in \mathcal{P}}U(g,P_2) = \overline{\int_a^b} f + \overline{\int_a^b} g $$