If it is correct to claim that the function $\frac{1}{\sqrt t}$ is not continuous?

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There is "Sufficient Condition Theorem" to test if a function's Laplace transform exists.

If a function $f$ is piecewise continuous on $[0, \infty)$ and is of exponential order $\alpha$, then the Laplace transform of $f$ exists for $s>\alpha$.

I read a document stating that this sufficient condition may not be necessary, i.e., a function which is not piecewise continuous on $[0,\infty)$ can also have its Laplace transform exist. Then take $f(t)=t^{-\frac12}$ for example.

However, $f(t)$ will have no definition at $t=0$ therefore, I am not completely convinced by the example.

Does it mean that all continous functions have to be defined in a closed domain?

Then what other examples can be more persuasive to the sufficient condition theorem?