Prove the maximum of two integrals of functions is less than or equal to the integral of the maximum of two functions

Question

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!

Answer 1

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*}