Question: Let $\{f_k(x)\}_{n=1}^\infty$ be a sequence of nonnegative functions on $\mathbb{R}$ such that $\sup_{x\in\mathbb{R}}f_k(x)=\frac{1}{k}$, and $\int_{\mathbb{R}}f_k(x)dx=1$. List all the possible values for $\int_{\mathbb{R}}\sup_{k\in\mathbb{N}}f_k(x)dx$.
My Thoughts: I am a bit confused on how we are going to relate the supremum over $x\in\mathbb{R}$ of $f_k(x)$, as in the hypothesis, and the supremum over $k\in\mathbb{N}$ of $f_k(x)$, as in what we are trying to prove... are they the same? Since the $f_k$'s are integrable, we can say they are measurable, so the Monotone Convergence Theorem would imply, I believe, that $\lim_{k\rightarrow\infty}\int_{\mathbb{R}}\sup_{x\in\mathbb{R}}f_k(x)dx=\int_\mathbb{R}\frac{1}{k}$ (maybe I can't put that supremum inside the integral like that....) But then I am not sure how to use $\int_{\mathbb{R}}f_k(x)dx=1$.... any help is greatly apprecaited! Thank you.
I think the answer is meant to be $\{\infty\}$. Suppose $$ \int_{\mathbb R} \sup_{k} f_k = I < \infty. $$ By the monotone convergence theorem, there exists a positive integer $n$ such that $$ \int_{[-n,n]} \sup_{k} f_k \ge I-\tfrac13. $$ Now consider $f_{6n}$. Since $\sup_{x\in\mathbb R} f_{6n}(x) \le \frac1{6n}$, we see that $\int_{[-n,n]} f_{6n} \le \frac13$. Therefore \begin{align} \int_{\mathbb R} \sup_{k} f_k &= \int_{[-n,n]} \sup_{k} f_k + \int_{\mathbb R \setminus [-n,n]} \sup_{k} f_k \\&\ge \int_{[-n,n]} \sup_{k} f_k + \int_{\mathbb R \setminus [-n,n]} f_{6n} \\&\ge I - \tfrac13 + \tfrac23 \\&= I + \tfrac13.\end{align}