# Definition of H(0) when inversing Laplace transform results in heaviside step function

We have: $$L^{-1}\{e^{-cs}F(s)\} = H(t - c)f(t-c)$$, with $$H$$ is a heaviside function.

In many documents, $$H(t - c)$$ is defined as:

$$H(t - c) = \left\{\begin{matrix} 0 &, t < c \\ 1 &, t \geq c \end{matrix}\right.$$

However, the Wikipedia seems to hint that the value of $$H(t - c)$$ at $$t = c$$ is actually of our choice. I know Wikipedia is not a reliable source but it is true that the definition of the heaviside function in the discrete form does vary at $$H(0)$$.

dlmf.nist.gov defines $$H(0)$$ to be $$0$$ whereas uea.ac.uk does not define $$H(0)$$.

So when we inverse the Laplace transform of a function and get the heaviside function as the result, we can define $$H(0)$$ at our choice?