Interchange between Integral and Sup?

Question

What is the analysis theorem that gives interchange between Integral and Sup. In particular, let $f_r (t)$ be a function on the circle $U = \{z\in \mathbb C: | z | = 1\}$, which condition must check $f_r (t)$ so that we can interchange Integral and sup: $$\sup_{0\leq r<1}\int_{\mathbb U}f_r(t) \, dt = \int_{\mathbb U}\, \sup_{0\leq r<1} f_r(t) \, dt\quad ?$$

Thank you in advance

Answer 1

It's more common (and equivalent) to work with infima (just put $-$ signs everywhere), so we want to find when $$ \inf_{r \in A} \int_X f_r = \int_X \inf_{r \in A} f_r. $$ Let $\inf_r f_r(t) = g(t)$. Then $\inf_r g(t) = g(t)$, and the right-hand side of the equality is $\int g$, so if we subtract and let $h_r=f_r-g$, equality occurs when $$ \inf_r \int_X h_r = 0 $$ But $f_r \geq \inf_r f_r = g $, so $f_r-g = h_r \geq 0$. So the integral is always nonnegative. The infimum is zero if and only if there is a sequence $r_n$ so that $\int h_{r_n} \to 0$. In turn, this means that it is possible to extract a subsequence $r_{n(k)}$ so that $h_{r_{n(k)}} \to 0$ almost everywhere (see e.g. here) (thanks to zhw. for reminding me that passing to a subsequence is necessary). Indeed, the subsequence can be chosen so that the convergence is uniform almost everywhere.

Therefore equality requires that there is a sequence, which we may call $f_{r_n}$ so that $f_{r_n} \to g$ uniformly almost everywhere. Is this sufficient? Yes, since in this case $\mu(X)<\infty$, so Hölder's inequality implies that $\int_X (f_{r_n}-g) < \lVert f_{r_n}-g \rVert_{\infty} \mu(X) $, so $f_{r_n} \to g$ uniformly almost everywhere if and only if $\int_X f_{r_n} \to \int_X g $.

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So a necessary and sufficient condition in your case is that there is a sequence $r_n$ so that $f_{r_n} \to \sup_r f_r$ uniformly.

Answer 2

We have from the property of $\sup$: $$\forall r\in[0,1):~~~f_r(t)=\sup_{0\leq r<1}f_r(t)$$ in your question you suppose that the supremum of integrable functions is integrable, then with integration over $U$ $$\forall r\in[0,1):~~~\int_{U}f_r(t)dt=\int_{U}\sup_{0\leq r<1}f_r(t)dt$$ this gives $$\sup_{0\leq r<1}\int_{U}f_r(t)dt=\int_{U}\sup_{0\leq r<1}f_r(t)dt$$ because $\int_{U}\sup_{0\leq r<1}f_r(t)dt$ is an upper bound for $\int_{U}f_r(t)dt$ where $0\leq r<1$.