I am going through the proof in Royden's Real Analysis of the following proposition:
Let f and g be be bounded measurable functions on a set of finite measure E. Then for any $\alpha$ and $\beta$:
$\int_{E}(\alpha f+\beta g)=\alpha\int_{E}f+\beta\int_{E}g$
The proof starts by establishing linearity for $\beta=0$, which I could follow without problems. It then goes into establishing linearity for $\alpha=\beta=1$. It goes like this: Let $\psi_{1}$ and $\psi_{2}$ be simple functions for which $f\leq\psi_{1}$ and $g\leq\psi_{2}$ on E. Then $\psi_{1}+\psi_{2}$ is a simple funciton and $f+g\leq\psi_{1}+\psi_{2}$ on E. Hence, since $\int_{E}(f+g)$ is equal to the Upper Lebesgue integral of f+g over E, by the linearity of integration for simple functions, $\int_{E}(f+g)\leq\int_{E}(\psi_{1}+\psi_{2})=\int_{E}\psi_{1}+\int_{E}\psi_{2}$
To which point I can follow and understand, however the next paragraph is the one I am having difficulties understanding:
The greatest lower bound for the sums of integrals on the right-hand side, as $\psi_{1}$ and $\psi_{2}$ vary among simple functions for which $f\leq\psi_{1}$ and $g\leq\psi_{2}$, equals $\int_{E}f+\int_{E}g$. These inequalities tell us that $\int_{E}(f+g)$ is a lower bound for these sums. Therefore, $\int_{E}(f+g)\leq\int_{E}f+\int_{E}g$
Specifically, how are we moving from
$inf\{\psi_{1}:\psi_{1}\mbox{ simple and }f\leq\psi_{1}\} = \int_{E}f$
$inf\{\psi_{2}:\psi_{2}\mbox{ simple and }g\leq\psi_{2}\} = \int_{E}g$
To the result?