"Almost surely" used in an expectation

Question

Let $(\mathsf{X}, \mathcal{X})$ be a measurable space, $\pi(dx)$ be a probability measure on it, and $K:X\times\mathcal{X}\to[0, 1]$ be a Markov kernel. I have the following property $$ \int K(x, A) \pi(dx) = 0 $$ In the textbook then they say that this means that $\pi$-almost surely $K(x, A) = 0$.

I don't understand why one can use almost surely here. My understanding is that almost surely can be used about events, but this is an expectation $\mathbb{E}[K(X, A)]$.

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Attempted Proof

I tried to reconciliate the two, starting from the second. I know that $\pi$-almost surely $K(x, A) = 0$ means $$ \int_{\{x\in\mathsf{X}\,:\, K(x, A) = 0\}} \pi(dx) = 1 $$ Since $\pi$ is a probability measure and $$ \mathsf{X} = \{x\in\mathsf{X}\,:\, K(x, A) = 0\} \cup \{x\in\mathsf{X}\,:\, K(x, A) \neq 0\} $$ I also know that the size of the complement set $$ \int_{\{x\in\mathsf{X}\,:\, K(x, A) \neq 0\}} \pi(dx) = 0 $$ Since $K$ is a Markov kernel, it must take non-negative values, so this means $$ \int_{\{x\in\mathsf{X}\,:\, K(x, A) > 0\}} \pi(dx) = 0 $$ which I can write slightly differently $$ \int_{\{x\in\mathsf{X}\,:\, K(x, A) \in (0, 1]\}} \pi(dx) = 0 $$ and again using the definition of preimage $$ \int_{K(\cdot, A)^{-1}((0, 1])} \pi(dx) = 0 $$ using now an indicator function $$ \int_{\mathsf{X}} 1_{K(\cdot, A)^{-1}(0, 1]}(x) \pi(dx) = 0 $$ and by definition of the indicator function, this is the same as $$ \int_{\mathsf{X}} 1_{(0, 1]}(K(x, A)) \pi(dx) = 0 $$ and I think I could write $$ \int_{\mathsf{X}} K(x, A) \pi(dx) = \int_{\mathsf{X}} 1_{\{0\}}(K(x, A)) \pi(dx) + \int_{\mathsf{X}} 1_{(0, 1]}(K(x, A)) \pi(dx) = 0 + 0 = 0 $$

Answer 1

In a measure space $(\Omega, \boldsymbol{\mathfrak{A}}, \mu) ,$ let $f$ be a nonnegative extended real-valued $\boldsymbol{\mathfrak{A}}$-measurable function on a domain set $D\in\boldsymbol{\mathfrak{A}}.$

Then $$\int_D f~\mathrm d\mu=0\implies f=0~~\mathrm{a.e.}\tag 1\label 1$$ on $D.$

This is easy to check: the result is trivial if $\mu(D) =0.$ Let $\mu(D) > 0.$ Suppose $\eqref 1$ isn't true. Then $\mu(E) > 0$ where $E:= \{D:f> 0\}.$ Now $E=\cup_{k\in \mathbb N} E_k, ~E_k:=\{D:f\geq \frac1k\}.$ As $E$ is not of measure $0, $ this means there exists $k'\in\mathbb N$ such that $\mu(E_{k'}) > 0$ as $E_k\uparrow E$ and $\mu(E) > 0$ due to the assumption.

Consider the nonnegative simple function $\varphi_0$ on $D$ defined as $\varphi_0:={k'}^{-1}\mathbf 1_{E_{k'}}+0\cdot\mathbf 1_{{E_{k'}}^\complement};$ it is clear that $\varphi_0\leq f. $ Therefore $$\int_D f~\mathrm d\mu=\sup_{0\leq ~\varphi~\leq f}\int_D\varphi~\mathrm d\mu\geq \int_D \varphi_0~\mathrm d\mu={k'}^{-1}\mu(E_{k'}) > 0,$$ a contradiction.

Now a Markov kernel $K(\cdot~, ~A) $ is nonnegative. You can use this result to reach the conclusion.

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Reference:

$\rm [I]$ Real Analysis: Theory of Measure and Integration, J. Yeh, World Scientific, $2014, $ p. $159, $ sec. $\S8.$