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Let $\mu(\cdot)$ be a probability measure on $\mathbb{R}^m$, so that $\int_{\mathbb{R}^m} \mu(dw) = 1$.

Consider a function $f:\mathbb{R}^n \times \mathbb{R}^m \rightarrow \mathbb{R}^n$ such that

$\forall w \in \mathbb{R}^m$ the map $x \mapsto f(x,w)$ is continuous;

$\forall x \in \mathbb{R}^n$ the map $w \mapsto f(x,w)$ is measurable.

For any continuous $\epsilon: \mathbb{R}^n \rightarrow \mathbb{R}_{\geq 0}$, define the set-valued map $ F: \mathbb{R}^n \times \mathbb{R}^m \rightrightarrows \mathbb{R}^n $ as

$$ F(x,w):= f( x + \epsilon(x)\overline{\mathbb{B}}, w ) + \epsilon(x) \overline{\mathbb{B}} $$

Prove that $\forall x \in \mathbb{R}^n$ the set-valued map $w \mapsto F(x,w)$ is measurable.

Note: Definition of measurability for a set-valued map.

A set-valued mapping $S: T \rightrightarrows \mathbb{R}^q $ is measurable if for every open set $\mathcal{O} \subset \mathbb{R}^q$ the set $S^{-1}(\mathcal{O}) \subset T $ is measurable, i.e. $S^{-1}(\mathcal{O}) \in \mathcal{A}$ (some $\sigma$-field of subsets of $T$). In particular, the set $dom S = S^{-1}(\mathbb{R}^q) $ must be measurable.

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Do you mean that a set-valued map is measurable if preimage of the singleton $S^{-1}(\{O\})$ is measurable for all open sets $O$? Because $S$ is set-valued, then $O$ is an element of its co-domain not a subset, and as far as I'm concerned we don't know a priori that $S$ would be a bijection. Also, what does $\bar{\mathbb{B}}$ represent? –  Thomas E. May 25 '12 at 7:01
    
$\overline{\mathbb{B}}$ represents the closed unit ball. For the definition of measurability of a set-valued map, you can refer to the book [Rockafellar, Wets - Variational Analysis] (math.washington.edu/~rtr/papers/rtr-VarAnalysis-RockWets.pdf) –  Adam May 25 '12 at 15:20

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