p-value subtlety: greater-equal vs. greater As I am reading Wassermann's book All of Statistics, I notice a fine subtlety in the definition of p-values, which I cannot make sense of. Informally, the Wassermann defines the p-value as 

[..] the probability (under $H_0$) of observing a value of the test statistic the same as or more extreme than what was actually observed.

Emphasis added. The same more formally (Theorem 10.12):

Suppose that the size $\alpha$ test is of the form
reject $H_0$ if and only if $T(X^n) \ge c_\alpha$.
Then,
$$\text{$p$-value} = \sup_{\theta\in\Theta_0} P_{\theta_0}[T(X^n) \ge T (x^n)]$$
where $x^n$ is the observed value of $X^n$. If $\Theta_0=\{\theta_0\}$ then
  $$\text{$p$-value} = P_{\theta_0}[T(X^n) \ge T (x^n)]$$

Furthermore, Wassermann defines the p-value of Pearson's $\chi^2$ test (and other tests analogously) as:
$$\text{$p$-value} = P[\chi^2_{k-1} > T].$$
The part I like to ask for clarification is the greater-equal ($\ge$) sign in the first and the greater ($>$) sign in the second definition. Why don't we write $\ge T$, which would match the first quotation of "the same as or more extreme?"
Is this sheer convenience so that we compute the p-value as $1-F(T)$? I notice that R also use the definition with the $>$ sign, e.g., in chisq.test.
 A: The first point of $\geq$ is that the hypothesis space is topologically closed within the whole parameter space. Without considering randomness, this can be a useful convention if you have some assertion about a converging sequence of parameters belonging to the hypothesis because then you would know that the limit does not suddenly belong to the alternative. 
Now considering the probability distributions, they are (usually) right-continuous. That means that the mapping of the closed hypothesis space to the $[0,1]$ interval is closed again. That's why confidence intervals are also closed by convention.
This enhances mathematics. Imagine, you would construct a confidence interval for the location parameter of an asymmetric probability distribution. There, you would have to trade the length to the upper tail for the length to the lower tail. The probability in both tails should sum up to $\alpha$. To have the CI as informative as possible, you would have to shorten the CI's length such that its coverage probability is still $\geq 1-\alpha$. This is a closed set. You can find an optimal solution there by some iterative algorithm, e.g. Banach's fixed point theorem. If it were an open set, you cannot do this.
A: "As or more extreme" is correct.
Formally, then, if the distribution is such that the probability of getting the test statistic itself is positive, that probability (and anything equally extreme, such as the corresponding value in the other tail) should be included in the p-value.
Of course, with a continuous statistic, that probability of exact equality is 0. It makes no difference if we say $>$ or $\geq$.
