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Let $m$ be a probability measure on $W \subseteq \mathbb{R}^p$, so that $m(W)=1$.

Find $m$, a locally-bounded function $f:\mathbb{R}^n \times \mathbb{R}^m \times \mathbb{R}^p \rightarrow \mathbb{R}^n$ and a locally-bounded, discontinuous function $y: \mathbb{R}^n \rightarrow \mathbb{R}^m$ such that

$$ \int_W | f(x,y(x),w) | m(dw) \ \leq \ \frac{3}{4}|x| $$

$$ \int_W \max_{\tilde{y} \in Y(x)} | f(x,\tilde{y},w) | m(dw) \ \geq \ |x| $$

for all $x \in \mathbb{R}^n$, where $Y(x) := \bigcap_{\epsilon > 0} \{ y(x+ \epsilon \mathbb{B}) \} $ is the set of all limit points of $y(x)$ (remind that $y$ is discontinuous).

Note. I have already found the following solution: $m({-0.6}) = m(0.6) = 1/2$, $f(x,y,w) := x(y+w)$, $y(x) := 0.6$ if $x \in \mathbb{Q}$, $-0.6$ otherwise. In fact, we have $\mathbb{E}[|f(x,y(x),w|] = 0.6 |x| \leq 3/4 \cdot |x|$, and $\mathbb{E}[ \max_{\tilde{y} \in Y(x)} |f(x,\tilde{y},w)| ] = 1.2 |x| > |x|$.

There can be a solution where $y$ that is not discontinuous everywhere?

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