When determining a test does it matter if $T\geq c$ or $T>c$? Let's imagine that we're testing some hypothesis and we have a statistic $T$, for which we can build the following test with dimension $\alpha=Pr\{T\geq c | H_0 \}=P\{T>c_* | H_0\}$, where $c_*=c + \text{constant}$ or $c_*=c$ depending on whether we're dealing with a discrete or continuous r.v.
Which inequality should we use? And which type would be easier to generalize if we had a mixed variable(with some jumps)? 
Any help would be appreciated.
 A: If you have a test statistic with a mixed distribution, I would have thought you would want to retain the flexibility to choose between the two inequalities 
Consider the following cumulative distribution function (black) for the test statistic $T$ under the null hypothesis



*

*If you want the critical region to correspond to J and above, then it does not matter whether you choose $T \gt c_3$ or $T \ge c_3$ as the distinction has probability zero 

*Similarly if you want the critical region to correspond to N and above, then it does not matter whether you choose $T \gt c_1$ or $T \ge c_1$

*If you want the critical region to correspond to K and above, then you should choose $T \gt c_2$ 

*By contrast if you want the critical region to correspond to M and above, then you should choose $T \ge c_2$

*The harder question is if you want the critical region to correspond to L and above, where you will not get exactly the $\alpha$ you are aiming for unless you use a random tie-splitter when $T=c_2$.  Assuming that is not desirable, it is typical practice to take the next possible lower $\alpha$, which would involve $T \gt c_2$ 
A mixed distribution for the test statistic may be rare.  In discrete and continuous cases you can choose either approach, though personally I have a poorly motivated habit of often choosing the strict inequality, as this is associated with my confidence intervals usually being presented as closed and so their complements being open   
