Using the mean value theorem to prove that a non-zero continuous function is only defined at one point when mulltiplied by the dirac delta function I have been following the course in Introduction to Probability, Statistics and Random Processes, by Hossien Pishro-Nik, and I was having some troubles understanding proof in chapter 4.3.2 Using the Delta Function. The lemma states that
$$\int_{-\infty}^{\infty} g(x) \delta(x-x_0) dx = g(x_0).$$
The proof states that, let $I$ bet the value of the above integral. Then, we have
$$I= \lim_{\alpha \rightarrow 0} \bigg[ \int_{-\infty}^{\infty} g(x) \delta_{\alpha} (x-x_0) dx \bigg]$$
$$=\lim_{\alpha \rightarrow 0} \bigg[ \int_{x_0-\frac{\alpha}{2}}^{x_0+\frac{\alpha}{2}} \frac{g(x)}{\alpha} dx \bigg].$$
By the mean value theorem in calculus, for any $α>0$, we have
$$\int_{x_0-\frac{\alpha}{2}}^{x_0+\frac{\alpha}{2}} \frac{g(x)}{\alpha} dx=\alpha \frac{g(x_{\alpha})}{\alpha}=g(x_{\alpha}),$$
for some $x_{\alpha} \in (x_0-\frac{\alpha}{2},x_0+\frac{\alpha}{2}).$ Thus, we have
$$I = \lim_{\alpha \rightarrow 0} g(x_{\alpha})=g(x_0).$$
Now, the lemma makes sense intuitively but I am unable to understand why
$$\int_{x_0-\frac{\alpha}{2}}^{x_0+\frac{\alpha}{2}} \frac{g(x)}{\alpha} dx=\alpha \frac{g(x_{\alpha})}{\alpha}=g(x_{\alpha}).$$
From my understanding of the mean value theorem, shouldn’t
$$\int_{x_0-\frac{\alpha}{2}}^{x_0+\frac{\alpha}{2}} \frac{g(x)}{\alpha} dx=\frac{1}{\alpha}\frac{G(x_{0}+\frac{\alpha}{2})-G(x_{0}-\frac{\alpha}{2})}{(x_{0}+\frac{\alpha}{2})-(x_{0}-\frac{\alpha}{2})}$$
$$=\frac{1}{\alpha}\frac{G(x_{0}+\frac{\alpha}{2})-G(x_{0}-\frac{\alpha}{2})}{\alpha}=\frac{G(x_{\alpha})}{\alpha^2},$$
where $G(x_{\alpha})$ is $G(x)$ defined in some range $x_{\alpha} \in (x_0-\frac{\alpha}{2},x_0+\frac{\alpha}{2}).$
As you can see I am unable to understand how the author managed to achieve the result he stated. Can you please help me understand how the author achieved the result?
 A: Full credit goes to PNDas for his comment.
The mean value theorem states that if a function $f(x)$ satisfies the following conditions

*

*$f(x)$ is continuous on the closed interval $[a,b]$

*$f(x)$ is differentiable on the open interval $(a,b)$
then there is a number c such that $a<c<b$ and
$$f'(c) = \frac{f(b) - f(a)}{b - a}$$
$${\implies}{f(b)-f(a)} = {f'(c)}{(b-a)},$$
where $f'$ is the differential of $f$.
Since
$$\int_{a}^{b}{f'(x)dx} = {f(b)-f(a)},$$
then application of the mean value theorem would imply that
$$\frac{1}{b-a}\int_{a}^{b}{f'(x)dx} = {f'(c)}$$
$${\implies}\int_{a}^{b}{f'(x)dx} = {f'(c)}{(b-a)},$$
where $a<c<b$.
Applying this to the problem in the question would produce
$$\int_{x_0-\frac{\alpha}{2}}^{x_0+\frac{\alpha}{2}}{\frac{g(x)}{\alpha}dx} = {\frac{g(x_{\alpha})}{\alpha}}{[(x_0+\frac{\alpha}{2})-(x_0-\frac{\alpha}{2})]}$$
$$= {\alpha}{\frac{g(x_{\alpha})}{\alpha}} = g(x_{\alpha}),$$
where $x_{\alpha}{\space}{\in}{\space}({x_0-\frac{\alpha}{2}}, {x_0+\frac{\alpha}{2}}).$
The definition of the mean value theorem was taken from Paul's Online Notes.
