Induced Measure Integral

Question

In my text, I have the following change of variables formulas:

A function $f: \Omega \mapsto \mathbb{R}$ is integrable with respect to the induced measure $\mu (T^{-1})$ iff $f(T)$ is integrable with respect to $\mu$.

In this case, it holds that $$ \int_{\Omega}f(T(\omega))\mu(d \omega) = \int_{\Omega_{1}} f(\omega_{1})\mu(T^{-1}) (d \omega_{1}) $$ $$\int_{A}f(T(\omega))\mu(d \omega) = \int_{T(A)}f(\omega_{1})\mu(T^{-1}(A_{1}))(d\omega_{1}),\,\text{for any}\,A_{1} \in \mathcal{F}_{1}\,\,\,\,\,(*) $$ (where $\mathcal{F}_{1}$ is a $\sigma$-field). ($A \in \mathcal{F})$

Now, when $T(\omega) = \varphi(\omega) = \sum_{i=1}^{\infty}x_{i}I_{A_{i}}(\omega)$ and $f(x) = x^{2}$, I have to verify that $(*)$ holds.

Plugging in everything that I know into $(*)$, I have for the LHS: $$\int_{A_{1}}\left[ \varphi(\omega)\right]^{2}\mu(d \omega) = \int_{A_{1}} \left[ \sum_{i=1}^{\infty} x_{i} I_{A_{i}}(\omega)\right]^{2} \mu(d \omega).$$

For the RHS, I have that $$\int_{\varphi(A_{1})} \omega_{1}^{2}\mu \left( \varphi^{-1}(A _{1})\right)d(\omega_{1}) = \int_{\sum_{i=1}^{\infty}x_{i}I_{A_{i}} (A_{1})} \omega_{1}^2 \mu(\varphi^{-1}(A_{1}))d(\omega_{1})$$. And since it seems to me that $\sum_{i=1}^{\infty}x_{i}I_{A_{i}}(A_{1}) = x_{1}$, I would imagine that this last integral would equal $$ = \int_{x_{1}}\omega_{1}^2 \mu(\varphi^{-1}(A_{1}))d(\omega_{1})$$

Now, then, my goal is to show that $$ \int_{A_{1}} \left[ \sum_{i=1}^{\infty} x_{i} I_{A_{i}}(\omega)\right]^{2} \mu(d \omega) = \int_{x_{1}}\omega_{1}^2 \mu(\varphi^{-1}(A_{1}))d(\omega_{1}) $$

But, how do I do that? For one thing, what is the preimage under $\varphi$ of $A_{1}$? And then, what is its $\mu-$measure? Also, how do I get the sets that we're integrating over to be the same thing?

I am very stuck and really need help finishing this proof. I thank you ahead of time for your time and patience!

Answer 1

You should probably get hold of another text. The formula $(*)$ does not make sense in general. Let me change the notation a bit and use indexes 1 and 2 for the respective spaces. Let $$T:(\Omega_1,{\cal B}_1) \rightarrow (\Omega_2,{\cal B}_2)$$ be a measurable map between spaces equipped with $\sigma$-algebras. By definition for any $A_2\in {\cal B}_2$ the set $T^{-1} A_2 \in {\cal B}_1$ is thus measurable. But nothing assures that $A_1\in {\cal B}_1$ implies $TA_1\in {\cal B}_2$. If e.g. $\Omega_1=\Omega_2$ but ${\cal B}_2 \subsetneq {\cal B}_1$ (strict subset), then the identity map is measurable but its inverse (also the identity) is not!

When $\mu_1$ is a measure on $(\Omega_1,{\cal B}_1)$ you get the unique push-forward measure by declaring: $$ \mu_2 (A_2) = \mu_1 (T^{-1} A_2) = \mu_1 \circ T^{-1} (A_2) , \;\;\; A_2\in {\cal B}_2$$ This may be written in terms if integrals (using indicator functions) as: $$ \int_{\Omega_2} {\bf 1}_{A_2} \;d\mu_2 = \int_{\Omega_1} {\bf 1}_{T^{-1}A_2} \;d\mu_1= \int_{\Omega_1} {\bf 1}_{A_2} \circ T \; d\mu_1,$$ where I used that $x\in T^{-1} A_2$ iff $Tx \in A_2$. Taking linear combinations you get for a step function $f:\Omega_2\rightarrow {\Bbb R}$ $$ \int_{\Omega_2} f \;d\mu_2 = \int_{\Omega_1} f \circ T \; d\mu_1.$$ Passing through Lebesgue monotone convergence for positive step functions you get the result for any integrable function $f\in L^1(\mu_2)$. If you first restrict your measure to a set $\Lambda_2\in {\cal B}_2$ then you get first: $$ \int_{\Lambda_2} {\bf 1}_{A_2} \;d\mu_2 = \int_{T^{-1}\Lambda_2} {\bf 1}_{T^{-1}A_2} \;d\mu_1= \int_{T^{-1}\Lambda_2} {\bf 1}_{A_2} \circ T \; d\mu_1$$ and then for any integrable $f$ you get (as a theorem) the correct version of $(*)$: $$ \int_{\Lambda_2} f \;d\mu_2 = \int_{T^{-1}\Lambda_2} f \circ T \; d\mu_1.$$

In the example it is possible to calculate the push-forward measure but you would actually never do so, as in general it is very complicated. Instead simply use the above theorem. It does, however, become easy when the measurable sets in your sum are disjoint. So let us assume this. Again the notation is not strikingly optimal when using $A_i$ for several purposes, so let us write: $$ \phi = \sum_i x_i {\bf 1}_{E_i} $$ with the countable family $(E_i)_{i\geq 1}$ being a disjoint family. Then $$ \mu_2(A) = \mu_1(T^{-1} A) = \sum_{i: x_i\in A} \mu_1(E_i)$$ The support of $\mu_2$ is in $S=\{y |\exists i : x_i=y\}$ and you get: $$ \int_{\Lambda_2} f\;d\mu_2 = \sum_{y\in \Lambda_2} y^2 \mu_2(\{y\}) = \sum_{i: x_i\in \Lambda_2} x_i^2 \mu_1(E_i) = \int_{T^{-1} \Lambda_2} (T(\omega))^2 d\mu_1(\omega)$$

Answer 2

sorry, I was wrong. as d.k.o. said, T should be injective. here is the corrected version.

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ok let me expand my comments. I insists that $$ \int_{A}f(T(\omega))\mu(d \omega) = \int_{T(A)}f(\omega_{1})(\mu(T^{-1}))(d\omega_{1}),\,\text{for any}\,A \in \sigma (T^{-1}B), \text{where}\ B\in \mathcal F_1\,\,\,\,\,(*) $$

Let $T(\omega) = \varphi(\omega) = \sum_{i=1}^{\infty}x_{i}I_{A_{i}}(\omega)$, where $\{A_i\}$ disjoint and compose $\Omega$. Moreover let $f(x)=x^2$. For this problem, we must find a new $\sigma-$algebra to make $T^{-1}B$ well defined, I consider $T^{-1}{x_i}=A_i$, and $A\in \sigma(A_1,A_2,...)$.

Then the l.h.s. is $$ \int_{A}f(T(\omega))\mu(d \omega)=\int_{A}\sum_{i=1} ^{\infty}x_i^2I_{A_i} \mu(d \omega)=\sum_{i=1}^{\infty}x_i^2\mu(A\cap A_i) $$

note that the transformation $T$ returns only countable values $x_1,x_2,...x_n,...$, when returing $x_i$, this means $A\cap A_i\neq \phi$. so the r.h.s. is $$ \int_{T(A)}f(\omega_{1})(\mu(T^{-1}))(d\omega_{1})=\sum _{i=1}^{\infty}\int_{A}x_i^2I_{\{A\cap A_i\neq \phi\}}d\mu $$

Answer 3

I think that the second equality may work only for injective mapping $T$. Here $\mu\circ T^{-1}$ is a discrete measure which is concentrated on $\{x_i\}$, i.e. $\mu\circ T^{-1}(B)=\sum_{x_i\in B}\mu(A_i)$ (assuming that the sets $A_i$ are disjoint and $\bigcup_{i}A_i=\Omega$). So

$$ \int x^21_{T(A)}(\mu\circ T^{-1})(dx)=\sum_{x_i\in T(A)} x_i^2\mu(A_i)\ge\sum x_i^2\mu(A\cap A_i) $$

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Here is a simple example. Let $\mu$ be the Lebesgue measure on $\Omega=\mathbb{R}_{+}$. Take $T(\omega)=\lfloor \omega\rfloor$. Then $\mu\circ T^{-1}$ is the counting measure on $\mathbb{Z}_{+}$.

If $A=(1,3/2)$, then $\int_A \lfloor \omega\rfloor^2d\omega=0.5$. But $T(A)=\{1\}$ so that the second integral equals $1$.