1. Applying Fubini's theorem

Hello.

I haven't used Fubini's theorem for ages, and now that I came across a use of it, I am confused.

According to my notes, the following is justified by Fubini's theorem:

$\int_0^t\left(\int_s^t df(u)\right)dg(s) = \int_0^t\left(\int_{(0,u)}dg(s)\right)df(u).$

In my case, the lower limit of the integral is not included in the set over which we integrate, while the upper limit is, so for instance

$\int_0^t dg(s) = \int_{(0,t]}dg(s).$

Fubini's theorem states that under certain conditions, we have

$\int_A\left(\int_B f(x,y) dy\right) dx = \int_B\left(\int_A f(x,y) dx\right) dy,$

for some given pair of measures.

In my case, the set $B$ varies with $s$, which is a variable of integration in the outer integral. Upon the changing of the order of integration, why do the limits of integration appear as they do? It is not clear to me how to apply Fubini, when the sets $A$ and $B$ vary with the integration variables.

2. Re: Applying Fubini's theorem

Originally Posted by HappyJoe
Hello.

I haven't used Fubini's theorem for ages, and now that I came across a use of it, I am confused.

According to my notes, the following is justified by Fubini's theorem:

$\int_0^t\left(\int_s^t df(u)\right)dg(s) = \int_0^t\left(\int_{(0,u)}dg(s)\right)df(u).$

In my case, the lower limit of the integral is not included in the set over which we integrate, while the upper limit is, so for instance

$\int_0^t dg(s) = \int_{(0,t]}dg(s).$

Fubini's theorem states that under certain conditions, we have

$\int_A\left(\int_B f(x,y) dy\right) dx = \int_B\left(\int_A f(x,y) dx\right) dy,$

for some given pair of measures.

In my case, the set $B$ varies with $s$, which is a variable of integration in the outer integral. Upon the changing of the order of integration, why do the limits of integration appear as they do? It is not clear to me how to apply Fubini, when the sets $A$ and $B$ vary with the integration variables.
You can't apply Fubini's theorem (or at least not the usual version you stated) since you technically don't have an integral $\displaystyle \int_{A\times B}f\; d\mu$ you really just have an integral $\displaystyle \int_A f\; d\mu$ where $f$ is an integral.

3. Re: Applying Fubini's theorem

Originally Posted by HappyJoe
Hello.

I haven't used Fubini's theorem for ages, and now that I came across a use of it, I am confused.

According to my notes, the following is justified by Fubini's theorem:

$\int_0^t\left(\int_s^t df(u)\right)dg(s) = \int_0^t\left(\int_{(0,u)}dg(s)\right)df(u).$

In my case, the lower limit of the integral is not included in the set over which we integrate, while the upper limit is, so for instance

$\int_0^t dg(s) = \int_{(0,t]}dg(s).$

Fubini's theorem states that under certain conditions, we have

$\int_A\left(\int_B f(x,y) dy\right) dx = \int_B\left(\int_A f(x,y) dx\right) dy,$

for some given pair of measures.

In my case, the set $B$ varies with $s$, which is a variable of integration in the outer integral. Upon the changing of the order of integration, why do the limits of integration appear as they do? It is not clear to me how to apply Fubini, when the sets $A$ and $B$ vary with the integration variables.
I'm going to treat $dg(s)$ as ordinary Lebesgue measure, ie. $dg(s)=ds$, but I don't think it's a substantial assumption.

First $s\in [0,t]$ and $u\in[s,t]\subset[0,t]$ and so

$\int_0^t \int_s^t duds = \int_0^t \int_0^t 1_{[s,t]}(u)duds = \int_0^t \int_0^t 1_{[s,t]}(u)dsdu$

but $\int_0^t 1_{[s,t]}(u)ds= \int_0^u ds$ (this is easily seen if you draw the graph of $1_{[s,t]}(u)$ as a function of $(s,u)$) and this is exactly your result.

4. Re: Applying Fubini's theorem

Originally Posted by Jose27
I'm going to treat $dg(s)$ as ordinary Lebesgue measure, ie. $dg(s)=ds$, but I don't think it's a substantial assumption.

First $s\in [0,t]$ and $u\in[s,t]\subset[0,t]$ and so

$\int_0^t \int_s^t duds = \int_0^t \int_0^t 1_{[s,t]}(u)duds = \int_0^t \int_0^t 1_{[s,t]}(u)dsdu$

but $\int_0^t 1_{[s,t]}(u)ds= \int_0^u ds$ (this is easily seen if you draw the graph of $1_{[s,t]}(u)$ as a function of $(s,u)$) and this is exactly your result.
This makes sense.