Given a $\sigma$-additive measure space $(S,\Sigma,\mu)$ and a linear operator $$ U: L^\infty \to L^\infty $$ where $L^\infty$ is the space of essentially bounded measurable functions.

Assume it is know that the operator fulfills the following three properties

  1. $ f\geq 0 \Rightarrow Uf \geq 0$,
  2. $UI = I$ where $I(x)=1$ for all $x \in S$,
  3. $ ||U||\leq 1$,

where $|| \cdot ||$ is the operator norm induced by $L^\infty$.

I would like to know if this implies the following property:

Given a family of disjoint sets $(A_i)_{i \in \mathbb{N}} \in \Sigma$ and define $B = \bigcup_{i \in \mathbb{N}} A_i$ and $B_n = \bigcup_{i =1}^n A_i$, does then the sequence converge $$ U 1_{B_n} \to U 1_B$$ pointwise $\mu$-almost surely?

I know that the sequence $( U 1_{B_n})_{n\in \mathbb{N}}$ converges $\mu$-almost surely, because from condition $1.$ it is clear that the sequence is monotonically increasing, and it is clear that the sequence is bounded $\mu$-almost surely, however I am not sure if it converges towards $1_B$. I wonder if the linearity and condition $2.$ and $3.$ imply this.

Does anyone know?

  • $\begingroup$ Are you assuming that $\mu(S)=1$? It isn't stated in the problem, but the terminology/notation (almost surely, $\mathsf 1_{B}$ instead of $\chi_B$, etc. suggests that you're working in a probability space. $\endgroup$ – Math1000 Oct 3 '15 at 12:18
  • $\begingroup$ @Math1000 No its not a probability space, but it is $\sigma$-additive, I forgot to mention that. Thanks for your note. $\endgroup$ – Adam Oct 3 '15 at 12:21
  • $\begingroup$ Also I'm assuming that the norm $\|\cdot\|$ is the $L^\infty$ norm? $\endgroup$ – Math1000 Oct 3 '15 at 12:26
  • $\begingroup$ @Math1000 it is the operator norm induced by $L^\infty$, I will also add this to the question. Thank you. $\endgroup$ – Adam Oct 3 '15 at 12:29

It does not follow that $U 1_{B_n} \to U 1_B$ pointwise $\mu$-almost everywhere. Consider $S = \mathbb{N},\Sigma = \mathcal{P}(\mathbb{N})$ and $\mu$ the counting measure (or any measure given by positive masses on every point, so $\mu$ could even be a probability measure).

Let $\mathscr{U}$ be a free ultrafilter on $\mathbb{N}$ and define $\lambda_{\mathscr{U}} \colon L^\infty(S,\Sigma,\mu) \to \mathbb{C}$ by

$$\lambda_{\mathscr{U}}(f) = \lim_{\mathscr{U}} f(n)$$

and $U(f) := \lambda_{\mathscr{U}}(f)\cdot I$. Then $\lambda_{\mathscr{U}}$ is a positive linear functional with norm $1$ and $\lambda_{\mathscr{U}}(I) = 1$, so $U$ satisfies the conditions.

But with $A_i = \{i\}$ we have

$$U(1_{B_n}) = 0$$

for all $n\in \mathbb{N}$. Similar constructions can be made on many measure spaces.

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