Can the supremum and (measure theoretic) integral be interchanged?

Question

Let $(\Omega,\mu)$ be a measure space and let $A$ be any index set. Suppose that we have a family $\{f_{\alpha}\}_{\alpha\in A}$ of integrable maps $\Omega\to[0,\infty)$ such that that $\sup_{\alpha\in A}f_{\alpha}$ (the pointwise supremum) is finite everywhere and is integrable. Is it true that $$\sup_{\alpha\in A}\int_{\Omega}f_{\alpha} \ \text{d}\mu=\int_{\Omega}\sup_{\alpha\in A}f_{\alpha} \ \text{d}\mu?$$ The inequality "$\leq$" is clear as $$\int_{\Omega}f_{\alpha_{0}} \ \text{d}\mu\leq\int_{\Omega}\sup_{\alpha\in A}f_{\alpha} \ \text{d}\mu$$ for all $\alpha_{0}\in A$. How about the inequality "$\geq$"?

Answer 1

No, let $\Omega$ be any subset of $\Bbb{R}$ with finite, positive Lebesgue measure, and let $A=\Omega$. For each $\alpha\in A$, let $f_{\alpha}(x)=1$ if $x=\alpha$ and $0$ otherwise. Then, \begin{align} \sup_{\alpha\in A}\int_{\Omega}f_{\alpha}\,d\lambda &= 0<\lambda(\Omega)=\int_{\Omega} 1\,d\lambda =\int_{\Omega}\sup_{\alpha\in A}f_{\alpha}\,d\lambda. \end{align}

Answer 2

It's not true even when $A$ is finite. Example: $f_1 = \chi_{[0,1]}$ and $f_2 = \chi_{[1,2]}$. Then

$$\sup \left\{ \int f_1, \int f_2\right\} = 1 < 2 = \int \sup \{ f_1, f_2\}.$$