Why is $f(x)=x^2$ not equal to zero a.e. in regards to Lebesgue measure?

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In our measure theory script (and in further literature also), to illustrate the term "almost everywhere" it's stated that the function $x^2$ is not equal to zero a.e. considering Lebesgue measure while it is equal to zero considering dirac measure in 0. I have difficulties in understanding this.

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$x^2$ is equal to $0 $ only at $x=0$ and one point has zero Lebesgue measure. However, this Dirac measure in $0$ of this one-point is $1$. This Dirac measure is defined as $$ \delta_0(A) = \begin{cases} 0, & x\notin A \\ 1, & x\in A \end{cases} $$ for any measurable set $A$. So, $\mu(\{x:x^2=0\}) = 0$ and $\mu(\{x:x^2\neq0\}) > 0$, where $\mu$ is Lebesgue measure, and $\delta_0(\{x:x^2=0\}) = 1$ and $\delta_0(\{x:x^2\neq0\}) = 0$. So, $x^2 \neq 0$ a.e. for Lebesgue measure, but not for Dirac measure.