Maximum of bounded expectations at a certain Borel set?

Question

Assume ${\bf x} \in \mathbb{R}^n$ denotes a real-valued bounded random variable with a distribution measured on the Borel space $(\mathbb{R}^n,\mathcal{B}^n)$. Let $f:\mathbb{R}^n\to\mathbb{R}$ denote a bounded Borel measureable function. Then, the following expectation value for any Borel set $B$

$$ E[f({\bf x}){\bf 1}_{{\bf x}\in B}] = \int_B f({\bf x}) dP $$

is bounded.

Is there any reason that the $\sup_{B\in\mathcal{B}^n} E[f({\bf x}){\bf 1}_{{\bf x}\in B}]$ occurs at a certain Borel set $B$ or could there just be convergence towards the supremum?

Answer 1

Yes, it's attained. Note that the desired expression can be written as $E[(f 1_B)(\mathbf{x})]$. Then it's clear that we get the maximum by taking $B = \{f \ge 0\}$, so that $f 1_B = f^+$, the positive part of $f$. Indeed, if $A$ is any other Borel set, then $E[(f 1_A)(\mathbf{x})] = E[(f^+ 1_A)(\mathbf{x})] - E[(f^- 1_A)(\mathbf{x})]$. But $f^+ 1_A \le f^+$ and $f^- 1_A \ge 0$.

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Previous overly sophisticated argument, please ignore:

Let $\nu(B) = E[f({\bf x}){\bf 1}_{{\bf x}\in B}] = \int_B f({\bf x}) dP$; then $\nu$ is a signed measure on $\mathbb{R}^n$. The Hahn decomposition theorem guarantees that we can partition $\mathbb{R}^n = B_+ \cup B_-$, with $B_+, B_-$ Borel, such that $\nu(A) \ge 0$ for all Borel $A \subset B_+$ and $\nu(A) \le 0$ for all Borel $A \subset B_-$. In particular, if $B$ is any other Borel set, we have $\nu(B) = \nu(B_+) - \nu(B_+ \setminus B) + \nu(B \cap B_-)$. But $\nu(B_+ \setminus B) \ge 0$ by the theorem, and $\nu(B \cap B_-) \le 0$, so $\nu(B) \le \nu(B_+)$. Thus the maximum is attained at $B_+$.