If the CDF is defined:
$F_X\left(x\right)=\:1-e^{-λx}\::\:x\ge 0\:and\:0:\:x<0$
But when we substitute $x=0$ in the equation $1-e^{-λx}$ we got $0$.
So why we can't say that:
$F_X\left(x\right)=\:1-e^{-λx}\::\:x>0\:and\:0:\:x\le 0$
CDF – Cumulative distribution function
conditional probabilitycumulative-distribution-functionsprobabilityprobability distributionsrandom variables
Best Answer
Both statements are true. The reason we usually write the first is that although the CDF is $0$ at $x=0$, the density isn't. The PDF is $f(x)=\lambda e^{-\lambda x}$, which is $\lambda$ at $x=0$.
Writing the CDF in the first way makes clearer the distinction between $x\geq0$, where the denisty is positive, and $x<0$, where the density is $0$.
As Ian points out, the above is probably not technically rigorous (although it is a good intuitive reason as to why we should lump $x=0$ together with the points $x>0$, rather than with the points $x<0$).
A more rigorous answer, as Stefan hints at, is that the set $x\geq0$ is the support of the distribution, i.e. the smallest closed set $S\subset\mathbb R$ such that $\mathbb P(X\in S)=1$. So it makes sense to define the CDF separately for two types of points in $\mathbb R$: those in the support of $X$ (i.e. $x\geq0$), and those not in the support of $X$ (i.e. $x<0$).