Can we say Dirac delta function is zero almost surely? It is known that $\delta(x) = \infty $ if $x = 0$ and $=0$ if $x\ne 0 $ and we also know that $\int_{-\infty}^{\infty}\delta(x)dx=1$.  However, if we consider a Lebesgue integration, $\delta(x)$ is zero almost surely so that we can get $\int_{-\infty}^{\infty}\delta(x)d\mu(x)=0$.
Why I get a contradiction here?
Many thanks!

Because that is a bad "definition" of the "Dirac function"!  The "Dirac function" is not a function at all, it is a "distribution" or "generalized function", a functional that assigns a number to every function.  Specifically, the "Dirac function" assigns the number f(0) to every function, f.
