never user this site before and here I go: this example got my mind out
$$\int_a^b f(x)dx = \int_a^b f(x)\cdot\left(\int_{a-\Delta a}^{b+\Delta b}\delta(x-y)dy\right)dx=*$$
$\text{Here }\Delta a, \Delta b > 0, \delta \text{ denotes delta function.}$ So it seems legal for me to assume that for any $x$ internal integral equals 1 and it's okay. Let's rewrite expression as follows: $$*=\int_a^bf(x)\cdot\left(\lim_{n\rightarrow\infty}\int_{a-\Delta a}^{b+\Delta b}\delta_n(x-y)dy\right)dx = \text{Fubini theorem works?}=\lim_{n\rightarrow\infty}\int_{a-\Delta a}^{b+\Delta b}dy\cdot\int_a^bf(x)\delta_n(x-y)dx=\int_{a-\Delta a}^{b+\Delta b}dy\cdot\lim_{n\rightarrow\infty}\int_a^bf(x)\delta_n(x-y)dx=\int_{a-\Delta a}^{b+\Delta b}f(y)dy$$ Here $\lim\limits_{n\rightarrow\infty}$ should be understanded as weak limit. So we have got that $\int\limits_a^b f(x)dx = \int\limits_{a-\Delta a}^{b+\Delta b}f(y)dy$ which is actually strange for me since I see no error. My thoughts is as follows:
- Weak convergence is seemingly Ok, as far as I can see
- Not sure about Fubini theorem since I never really used it directly. The definition I found on Wikipedia states that integrand has to be integrable and measurable in the given domain. It is integrable since I state $f(x)$ is integrable under single integral and so $f(x)\delta(x-y)$ does under double integral (is that right?). It is measurable at least for $\delta_n$. So Fubini theorem seemingly works?
- Maybe last expression can be in some way interpreted as a correct one?
I really want to know whether there is some possibility to deal with this example properly. Thanks in advance for any tryings.
P.S. yes I am physicist and my math skill is not really good