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Let $\mu(\cdot)$ be a probability measure in $X$.

Consider a function $f: Z \times X \rightarrow \mathbb{R}_{\geq 0}$, with $Z \subset \mathbb{R}^n$ compact, such that:

  1. $\forall x \in X, \quad z \mapsto f(z,x)$ is continuous;

  2. $\forall z \in Z, \quad x \mapsto f(z,x)$ is measurable;

The family $\mathcal{F} = \{ f(z,\cdot) \}_{z \in Z}$ is a subset of $L^1(\mu)$:

\max_{z \in Z} \int_X f(z,x) \mu(dx) < \infty

Is the closure of $\mathcal{F}$ in the weak topology of $L^1(\mu)$ compact?

2 Answers 2

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No.

Take $X = [0,1]$, $\mu$ Lebesgue measure, $n=1$, $Z= \{\frac{1}{n} : n \in \mathbb{N}\} \cup \{0\}$. Set $f(\frac{1}{n}, x) = n 1_{(0, \frac{1}{n})}(x)$, $f(0,x) = 0$. It is easy to check $f$ satisfies the given conditions, but $\{f(z,\cdot) : z \in Z\}$ is not weakly precompact in $L^1$.

Intuitively, the weak limit is trying to be a point mass at 0, which is not in $L^1$. To prove this more carefully, let $g$ be a weak limit point of $\{f(z,\cdot): z \in Z\}$. and consider what $\int_\epsilon^1 g$ must be for any $\epsilon > 0$.

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Take $\mu = $ normalized lebesgue measure on $[-1,1]$. Slightly alter a family of approximate identities to satisfy 1. Let $f >0, \int f(x)dx = 1, f $ continuous except at $0$ where $f(0) = 0$, and $f(z) = \frac {f((1-z)t)} {(1-z)}$. Then any weak limit of this sequence is a point mass at zero.