Norris Markov Chains expected hitting time formula

Question

I am currently working through Norris' Markov Chains book, but I am a bit confused about the introduction of the expected hitting time.

We define the hitting time of a subset $A$ of the state space $I$ as the random variable $H^A: \Omega \to \{0,1,2,...\} \cup \{\infty\}$ where $$H^A = \text{inf}\{n \geq 0 | X_n(\omega) \in A\}$$ and the infimum of the empty set is said to be $\infty$.

Following this, we are told: "The mean time taken for $(X_n)_{n\geq 0}$ to reach $A$ is given by $$\mathbb{E}_i(H^A) = \sum_{n \lt \infty} n\mathbb{P}(H^A = n) + \infty\mathbb{P}(H^A = \infty)."$$

I understand that this is just the definition of expectation, with the infinity term explicitly separated from the main sum. My question is, how is it valid to write the final term? I assume the interpretation here is that provided it is possible to hit $A$ in this Markov Chain, $\mathbb{P}(H^A = \infty) = 0$ by the definition of $H^A$, so the infinity and the 0 cancel each other out (and take a value of 1?). Otherwise if it is impossible to hit $A$ the expected time to hit is $\infty$ which makes sense intuitively.

The problem I have here is that this just all seems like an abuse of notation. Is it really ok to just write $\infty$ in a product like that? And if so, is my assumption about its interaction with being multiplied by 0 correct?

Answer 1

In measure theory it is a popular convention to define $0\cdot \infty = 0$, which helps to compactly write formulas such as the one above.

I consider the definition above as a shorthand to

The mean time to reach $A$ is defined as $\infty$ if $\mathbb P(H^A=\infty) > 0$ and as $\sum_{n<\infty} n\, \mathbb P(H^A=n)$ otherwise.