I have to show that $U(f,R) = U(f,P) + U(f,Q)$ where $R = P \cup Q$, I found a way where I proved that $U(f,R) - U(f,P) = U(f,Q)$, but it's sort of long and doesn;t translate well to prove the same result but for $R = P_1 \cup P_2 \cup P_3 \cup ... \cup P_n$. Is there a better, more simpler way to prove it?
2026-03-31 22:47:49.1774997269
Showing that $U(f,R) = U(f,P) + U(f,Q)$ where $R = P \cup Q$
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Since $R$ is a refinement of both $P$ and $Q$, the upper sum for $R$ is less than or equal to each of the upper sums for $P$ and $Q$. Hence, this is not true in general.
For example, if $f$ is positive then
$$0 < U(R,f) \leqslant U(P,f), \,\,\,\,0 < U(R,f) \leqslant U(Q,f), $$
and $U(R,f) < U(P,f) + U(Q,f)$.
An exception where it holds is the trivial case where $f(x)$ is everywhere $0$, and perhaps some contrived partitions when $f(x)$ is not always of the same sign.