Let $\mu_1,\ldots\mu_n$ be $\sigma$-finite measures and $h_i$ probability densites on $(\Omega_i,\mathcal{A}_i,\mu_i),~i=1,\ldots,n$. Show that for $$ h(\omega_1,\ldots\omega_n):=\prod_{i=1}^{n}h_i(\omega_i) $$ it is $$ (h_1\mu_1)\otimes\ldots\otimes (h_n\mu_n)=h(\mu_1\otimes\ldots\otimes\mu_n) $$ and that this defines a probability measure.

Good evening,

The first thing I showed was that the measures $h_i\mu_i, i=1,\ldots,n$ are $\sigma$-finite measures on $(\Omega_i,\mathcal{A}_i), i=1,\ldots,n$, i.e. it makes sense to talk about the product measure of the measures $h_i\mu_i, i=1,\ldots,n$. (Is this a $\sigma$-finite measure?)

Starting on the LHS, I get the following:

For $A:=A_1\times\ldots\times A_n\in\bigotimes_{i=1}^{n}\mathcal{A}_i$ it is $$ (h_1\mu_1)\otimes\ldots\otimes (h_n\mu_n)(A)=\int_{\Omega_1}\ldots\int_{\Omega_n}(1_A)_{\omega_1\times\ldots\times\omega_{n-1}}(\omega_n)\, dh_n\mu_n\ldots\, dh_1\mu_1\\=\int_{\Omega_1}\ldots\int_{\Omega_n}1_{A_1}(\omega_1)1_{A_2}(\omega_2)\cdot\ldots\cdot1_{A_n}(\omega_n)\, dh_n\mu_n\ldots\, dh_1\mu_1\\=\int_{A_1}\int_{A_2}\ldots\int_{A_n}\, dh_n\mu_n\ldots\, dh_2\mu_2\, dh_1\mu_1\\=\int_{A_1}\int_{A_2}\ldots\underbrace{\int_{A_n}h_n\, d\mu_n}_{=h_n\mu_n(A_n)}\ldots\, dh_2\mu_2\, dh_1\mu_1\\=h_n\mu_n(A_n)\int_{A_1}\ldots\int_{A_{n-1}}h_{n-1}\, d\mu_{n-1}\ldots \,dh_1\mu_1\\=h_{n-1}\mu_{n-1}(A_{n-1})h_n\mu_n(A_n)\int_{A_1}\ldots\\=\ldots\\=h_1\mu_1(A_1)\cdot h_2\mu_2(A_2)\cdot\ldots\cdot h_n\mu_n(A_n)\\=\prod_{i=1}^{n}h_i\mu_i(A_i) $$

Starting on the RHS, I get for $A$: $$ h(\mu_1\otimes\ldots\otimes\mu_n)(A)=\int_A h\, d(\mu_1\otimes\mu_2\ldots\otimes\mu_n). $$ Because the $\mu_i, i=1,\ldots,n$ are $\sigma$-finite, $h$ is measurable (because $h_i, i=1,\ldots,h_n$ are measurable) and $h$ is non-negative (because $h_i, i=1,\ldots,n$ are non-negative), one can apply Fubini: $$ h(\mu_1\otimes\ldots\otimes\mu_n)(A)=\int_A h\, d(\mu_1\otimes\mu_2\ldots\otimes\mu_n)\\=\int_{A_1}\ldots\int_{A_n}h_{\omega_1\times\ldots\times\omega_{n-1}}(\omega_n)\, d\mu_n(\omega_n)\ldots\, d\mu_1(\omega_1)\\=\int_{A_1}\ldots\int_{A_n}h_1(\omega_1)\cdot h_2(\omega_2)\cdot\ldots\cdot h_n(\omega_n)\, d\mu_n(\omega_n)\ldots\, d\mu_1(\omega_1)\\=\prod_{i=1}^{n}h_i\mu_i(A_i) $$ So it is LHS=RHS. Because $A$ was arbitrarily chosen out of $\bigotimes_{i=1}^{n}\mathcal{A}_i$, this identity holds for any $A\in\bigotimes_{i=1}^{n}\mathcal{A}_i$. So both measures (LHS and RHS) are identitical.

Now it remains to show that this defines a probability measure. From the shown identity it follows that it does not matter which measure (LHS or RHS) I choose to show that.

I choose the right side.

If I see it right, the only thing to show is that $$ h(\mu_1\otimes\ldots\otimes\mu_n)(\times_{i=1}^{n}\Omega_i)=1. $$ But this follows from the fact that $h_i, i=1,\ldots,n$ is a probability density and so $$ h_i\mu_i(\Omega_i)=\int_{\Omega_i}h\, d\mu_i=1, i=1,\ldots,n. $$ From this it follows with the result above $$ h(\mu_1\otimes\mu_1\otimes\ldots\otimes\mu_n)(\times_{i=1}^{n}\Omega_i)=\prod_{i=1}^{n}h_i\mu_i(\Omega_i)=\prod_{i=1}^{n}1=1. $$


Is my proof right?

Or is there anything additional to show in order to prove that it is a probability measure?

With greetings


  • 1
    $\begingroup$ it looks correct to me $\endgroup$ – i like xkcd Jan 18 '14 at 20:23
  • 1
    $\begingroup$ "Because $A$ was arbitrarily chosen out of $\bigotimes\limits_{i=1}^{n}\mathcal{A}_i$" Actually $A$ was chosen in $\prod\limits_{i=1}^{n}\mathcal{A}_i$, which is a much smaller collection of sets--hence indeed some steps are missing to show the measures are equal. $\endgroup$ – Did Jan 18 '14 at 21:41
  • $\begingroup$ We had in our reading, that the product measure is determined uniquely by all $A_1\times \ldots\times A_n\in\mathfrak{Z}(\mathcal{A}_i: i=1,\ldots,n)$. $\endgroup$ – math12 Jan 18 '14 at 21:46
  • $\begingroup$ I forgot to mention what $\mathfrak{Z}(\mathcal{A}_i: i=1,\ldots,n)$ is! Let $T$ be any index set and $\mathcal{E}(T)$ the set of the finite subsets of $T$, then $\mathfrak{Z}:=\mathfrak{Z}(\mathcal{A}_t: t\in T):=\bigcup_{S\in\mathcal{E}(T)}\left\{\prod_{t\in T} A_t: A_t\in\mathcal{A}_t, t\in S; A_t=\Omega_t, t\in T\setminus S\right\}$. $\endgroup$ – math12 Jan 19 '14 at 10:39
  • $\begingroup$ Our lecture says that a product measure is determined uniquely by all $A_1\times\cdots\times A_n\in\mathfrak{Z}$. So the measure on the LHS is. And because the measure on the RHS is the same for these $A_1\times\cdots\times A_n\in\mathfrak{Z}$, the two measures are identical. Right? $\endgroup$ – math12 Jan 19 '14 at 10:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.