Why $\lim_{\varepsilon \to 0}\int _{[0,1]}\delta _\varepsilon =1/2$ and not $1$ if $\delta (A)=\lim_{\varepsilon \to 0}\int_A \delta _\varepsilon $?

Question

I asked here a question a Dirac $\delta $ function, and Surb in Hamza Boulahia answers says that $$\delta (A)=\lim_{\varepsilon\to 0 }\int_A \delta _\varepsilon (x)dx,$$ for $$\delta _\varepsilon (x)=\frac{1}{\varepsilon }\boldsymbol 1_{[-\frac{\varepsilon }{2},\frac{\varepsilon }{2}]}(x).$$ If this is true, then for example $$\lim_{\varepsilon \to 0}\int_{[0,1]}\delta _\varepsilon (x)dx=1$$ since $\delta ([0,1])=1$. But when I do my calculation, I get $$\lim_{\varepsilon \to 0}\int_{[0,1]}\delta _\varepsilon =\lim_{\varepsilon \to 0}\int_0^{\frac{\varepsilon }{2}}\frac{1}{\varepsilon }dx=\frac{1}{2}\neq 1,$$ so is this definition for $\delta $ function wrong ?

Answer 1

It is indeed wrong to assert $\delta(A)=\lim_{\varepsilon\to 0}\int_A\delta_\varepsilon$. The $\int_{\mathbb{R}} f\,\mathrm{d}\delta=\lim_{\varepsilon\to 0}\int_{\mathbb{R}}f\delta_\varepsilon$ pairing is for $f$ continuous, which your $\mathbf{1}_{[0,1]}$ fails.

Alternatively, view $\delta_\varepsilon$ as a sequence of probability measures and we take $\delta$ to be the distribution limit. We have $\delta(A)=\lim_{\varepsilon\to 0}\delta_\varepsilon(A)$ for any set $A$ such that $\delta(\partial A)=0$ by Portmentau's theorem, but obviously $\delta(\partial[0,1])=\delta(\{0,1\})=1$.