TITLE: Prove the maximum of two integrals of functions is less than or equal to the integral of the maximum of two functions
QUESTION [6 upvotes]: Prove that on an interval $[a,b]$, if $f$ and $g$ are integrable, than $\max(f,g)$ is integrable and 
$$\max(\int_a^b f(x)dx, \int_a^b g(x)dx) \leq \int_a^b \max(f(x), g(x))dx. $$
I proved the first part by
$$\max(a,b)=\frac{a+b+|a-b|}{2}$$
so
$$\max(f(x),g(x))=\frac{f(x)+g(x)+|f(x)-g(x)|}{2}$$
The absolute value, the sum, and the difference, and the multiplication by a constant of Riemann integrable functions are all Riemann integrable, hence the max is also integrable.
Can anyone help with the second part? Many thanks!

REPLY [3 votes]: We have 
\begin{align*}
f\leq f\vee g,
\end{align*}
so by taking integrals both sides, we have 
\begin{align*}
\int f\leq\int f\vee g,
\end{align*}
similarly,
\begin{align*}
\int g\leq\int f\vee g,
\end{align*}
so
\begin{align*}
\int f\vee\int g\leq\int f\vee g.
\end{align*}