An Inequality for Lower Riemann Sums I'm trying to prove that for any bounded functions $f,g : [a,b] \to \mathbb{R}$ the lower Riemann integrals obey the inequality
$$\underline{\int}_a^b(f(x)+g(x))dx \leq \underline{\int}_a^bf(x)dx + \underline{\int}_a^bg(x)dx $$
Using $\inf f(x) + \inf g(x) \leq \inf (f(x) + g(x))$, I was able to show that for lower sums with a partition $P$
$$ L(P,f) + L(P,g)\leq L(P,f+g)$$
I know that the lower integral is the supremum of lower sums for all partitions so
$$\sup_{P}(L(P,f) + L(P,g))\leq \sup_{P}L(P,f+g)=\underline{\int}_a^b(f(x)+g(x))dx $$
Also since $L(P,f) + L(P,g) \leq \sup_{P}L(P,f) + \sup_{P}L(P,g)$ we know that
$$\sup_{P}(L(P,f) + L(P,g))\leq \sup_{P}L(P,f) + \sup_{P}L(P,g)= \underline{\int}_a^bf(x)dx + \underline{\int}_a^bg(x)dx$$
But this does not help me show the desired inequality.
Can anyone please help me continue?
 A: Your inequality for the lower sums is correct.  However, your upper bounds, while both correct, do not relate the lower integrals.  
Assume on the contrary that
$$\underline{\int}_a^b [f(x)+g(x)] \, dx  <  \underline{\int}_a^b f(x) \, dx  +  \underline{\int}_a^b g(x) \,dx .$$
Then
$$\underline{\int}_a^b [f(x)+g(x)] \, dx  -  \underline{\int}_a^b g(x) \,dx  <  \underline{\int}_a^b f(x) \, dx,$$
and there exists a partition $P$ such that
$$\underline{\int}_a^b [f(x)+g(x)] \, dx  -   \underline{\int}_a^b g(x) \,dx  <  L(P,f)  \leqslant \underline{\int}_a^b f(x) \, dx.$$
Hence,
$$\underline{\int}_a^b [f(x)+g(x)] \, dx  - L(P,f)  <   \underline{\int}_a^b g(x) \, dx,$$
and there exists a partition $P’$ such that
$$\underline{\int}_a^b [f(x)+g(x)] \, dx  - L(P,f)  <  L(P’,g) \leqslant \underline{\int}_a^b g(x) \, dx,$$
and
$$\underline{\int}_a^b [f(x)+g(x)] \, dx   <  L(P,f) + L(P’,g) .$$
Now 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 
$$L(Q,f+g) \leqslant \underline{\int}_a^b [f(x)+g(x)] \, dx   <  L(P,f) + L(P’,g)  \leqslant L(Q,f) + L(Q,g).$$
This contradicts the inequality for lower sums, and, therefore
$$\underline{\int}_a^b f(x) \, dx + \underline{\int}_a^b g(x) \, dx \leqslant \underline{\int}_a^b [f(x) + g(x)] \, dx. $$
Note that if $f$ and $g$ were Riemann integrable, you would have a strict equality.  
A: This is basically the same proof as @RRL's but in a direct manner. 
Proof:
Consider any two partitions $P_1, P_2$ of $[a,b]$; let $P:=P_1 \cup P_2$. Since $P$ is a refinement of both $P_1$ and $P_2$, we have $L(P_1, f) \leq L(P, f)$ and $L(P_2, g) \leq L(P, g)$,
so 
$$
L(P_1, f) + L(P_2, g) \leq L(P, f) + L(P, g) \leq L(P, f+g) \leq \underline{\int} (f+g)
$$
Since $\forall P_1$, 
$$
L(P_1, f) \leq \underline{\int} (f+g) - L(P_2, g)
$$
taking $\sup$ over $P_1$ gives
$$
\underline{\int} f \leq \underline{\int} (f+g) - L(P_2, g)
$$
Similarly, since $\forall P_2$,
$$
L(P_2, g) \leq \underline{\int} (f+g) - \underline{\int} f
$$
taking $\sup$ again gives
$$
\underline{\int} g \leq \underline{\int} (f+g) - \underline{\int} f
$$,
i.e., the desired inequality:
$$
\underline{\int} (f+g) \geq  \underline{\int} f + \underline{\int} g
$$
Note that the key to the above $\sup$ manipulations was considering two lower Darboux sums defined on two different partitions, $P_1$ and $P_2$.
