$-\int_a^b F_{\mu}(x)\phi'(x)dx=\ldots=\int_a^b\phi(x)d\mu(x)$?

Asked May 23 '21 at 01:06

Active May 23 '21 at 11:35

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If I consider the interval $[a,b]$ and a positive measure $\mu$ on this interval, we can define a function $F_{\mu}(x)=\mu([a,x])$. I want to show that the distributional derivative of this function is exactly the measure $\mu$ (I am not sure if this is the case though). So I think I have to take a test function $\phi$ and compute $$\langle DF_{\mu},\phi\rangle=-\int_a^b F_{\mu}(x)\phi'(x)dx=-\int_a^b\mu([a,x])\phi'(x)dx$$

but I am not sure how to continue. It would make sense if I arrive at the end at $\ldots=\int_a^b\phi(x)d\mu(x)$.

edited May 23 '21 at 11:35

asked May 23 '21 at 01:06

edamondo

1,277

Why use $-\infty$ rather than $a?$ – Thomas Andrews May 23 '21 at 01:09
You probably want $\mu$ a finite positive measure. – Thomas Andrews May 23 '21 at 01:15
This follows from Lebesgue's integration by parts. Consider the measure $\mu_F$ on $(a,b]$ defined as $\mu_F(c,d]=F(d)-F(c)$ for all $a\leq c<d\leq b$ and $\nu(c,d])=\int^d_c\phi'(t),dt$. Let $G_\nu(x)=\nu(a,x]$. Then $\int^b_a F_.mu(x),G(dx)=F(b)G(b)-F(a)G(a)-\int^a_bG(x-),F(dx)$. Notice that $\nu$ is not necessarily a positive measure, but it it certainly a signed measure with total finite variation. – Mittens May 23 '21 at 02:11
@ThomasAndrews, thank you this was a typo – edamondo May 23 '21 at 11:30
@OliverDiaz, is $\int_a^b Fmu(x)G(dx)$ the same as $\int_a^bF_{\mu}(x)\mu_G(dx)$? – edamondo May 23 '21 at 11:37
@edamondo: yes. Let me know if you have further questions. – Mittens May 23 '21 at 14:03

$-\int_a^b F_{\mu}(x)\phi'(x)dx=\ldots=\int_a^b\phi(x)d\mu(x)$?

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