When we can change maximization with integration?

Question

Under what conditions does the following equation hold?

$\int {\mathop {\max }\limits_{g \in G} g\left( x \right)dx} = \mathop {\max }\limits_{g \in G} \int {g\left( x \right)dx} $

Note that $G$ is set of functions.

Answer 1

Let's use $\sup$ instead of $\max$ to allow a more general situation, but the results wouldn't change.

For any specific $\hat g \in G$ $$\hat{g}(x) \le \sup_{g\in G} g(x),$$ and so over any (measurable at least) set $A$, $$\int_A\hat{g}(x) dx \le \int_A \sup_{g\in G}g(x) dx.$$ And since this is true for every $\hat{g} \in G$, it is also true that $$\sup_{\hat{g}\in G}\int_A\hat{g}(x) dx \le \int_A \sup_{g\in G}g(x) dx.$$

(And we can take off our hats now.) (?) $$\sup_{g\in G}\int_A g(x) dx \le \int_A \sup_{g\in G}g(x) dx.$$

Equality does not hold in many cases. For instance, if $G=\{g_1,g_2\}$, where $$g_1 \colon [0,1] \to [0,1], \; g_1(x)=x$$ and $$g_2 \colon [0,1] \to [0,1], \; g_2(x)=1-x$$

Both integrals equal $1/2$ so $$\sup_{g\in G}\int_0^1 g(x) dx =1/2$$

But $$\sup_{g\in G}g(x)= \begin{cases} 1-x & x \in [0,1/2) \\ x & x \in [1/2,1] \\ \end{cases},$$ and so $$\int_0^1\sup_{g\in G}g(x)= 3/4.$$

Of course there are situations in which equality holds ($G$ has only one element; one element of $G$ dominates all the others;...), but I cannot think of a relevant one...