Marginally continuous measures

Question

Consider a continuous density f on $\mathbb{R}^{2}$, and suppose that $\mu$ is the corresponding Lebesgue-Stieltjes measure on the product space $\left(\mathbb{R}^{2},\,\mathcal{B}\left(\mathbb{R}^{2}\right)\right)$. Suppose that the respective marginals are given by $\mu_{1}$, and $\mu_{2}$, also representing the continuous marginal densities of f. Is it true that the measure $\mu$ is absolutely continuous with respect to the product measure $\mu_{1}\times\mu_{2}$? I've been able to work out that this is true on cylinder sets, and have seen it stated as being true elsewhere without details, but cannot work it out completely. Any help would be appreciated. Thank you.

Answer 1

The continuity of the density $f$ is unnecessary. Let $A$ be a set with $\mu_1\times\mu_2(A)=0$, as before. Let $f_1$ and $f_2$ denote the marginal densities of $f$ (e.g, $f_1(x) =\int f(x,y)\,dy$), let $A_1=\{x: f_1(x)=0\}$, $A_2=\{y:f_2(y)=0\}$, and $G=\{(x,y):f_1(x)f_2(y)=0\}$. PhoemueX has argued that $A\setminus G$ has Lebesgue measure $0$; consequently $\mu(A\setminus G)=0$. Clearly $A$ is a subset of the union of $A_1\times\mathbb{R}$, $\mathbb{R}\times A_2$, and $A\setminus G$. And, by the Fubini/Tonelli theorem, $$ \mu(A_1\times\mathbb{R})=\int_{A_1}\left[\int_{\mathbb{R}} f(x,y)\,dy\right]\,dx =\int_{A_1} f_1(x)\,dx =0. $$ Similarly, $\mu(\mathbb{R}\times A_2)=0$. It follows from subadditivity that $\mu(A)=0$. This proves that $\mu$ is absolutely continuous with respect to $\mu_1\times\mu_2$.

Answer 2

Let us assume that $\mu_1 \times \mu_2 (A)=0$. This means

$$ 0=\int \chi_A (x,y) \int f(x,z)\, dz \int f(w,y)\, dw \, d(x,y). $$

As the integrand is nonnegative, this implies that for Lebesgue almost all $(x,y)\in A$, we have

$$ \int f(x,z)\, dz \int f(w,y)\, dw =0. $$

But this implies either

$$ \int f(x,z)\, dz =0, $$ or $$ \int f(w,y)\, dw=0. $$ The first case yields $f(x,z)=0$ for almost all $z$, because the integrand is nonnegative. But by continuity of $f$, this implies $f(x,z)=0$ for all $z$ and in particular $f(x,y)=0$.

In the other case, a similar argument yields $f(x,y)=0$. As this holds for Lebesgue almost all $(x,y)$, we get

$$ \mu (A)=\int \chi_A (x,y) f(x,y)\, d(x,y)=0. $$ This completes the proof.

Remark: We did only use continuity of $f$ in each variable separately here.

Answer 3

1. Suppose $(S,\mathscr{S})$ and $(T,\mathscr{T})$ be measurable spaces, and let $\nu$ be a $\sigma$-finite measure on the product measurable space $(S\times T,\mathscr{S}\otimes\mathscr{T})$.
2. Assume that there are $\sigma$-finite measures $m_S$ and $m_T$ on $(S,\mathscr{S})$ and $(T,\mathscr{T})$ respectively such that $\nu\ll m_S\otimes m_T$, that is, there is an nonnegative $\mathscr{S}\otimes\mathscr{T}$-measurable function $f$ such that $$\nu(ds,dt)=f(s, t)\,m_S(ds)m_T(dt)$$ Then, marginals $\nu_S$ and $\nu_T$ of $\nu$ on $(S,\mathscr{S})$ and $(T,\mathscr{T})$ are given by \begin{align} \nu_S(A)&=\int_S\mathbb{1}_A(s)\Big(\int_T f(s, t)\,m_T(dt)\Big)\,m_S(ds)\\ \nu_T(B)&=\int_S\mathbb{1}_B(t)\Big(\int_S f(s, t)\,m_S(ds)\Big)\,m_T(dt) \end{align} and so, $\nu_S\ll m_S$, $\nu_T\ell m_T$ and $$ \frac{d\nu_S}{dm_S}(s)=\int_T f(s, t)\,m_T(dt),\qquad \frac{d\nu_T}{dm_T}(t)=\int_S f(s, t)\,m_S(ds)$$

If $\nu_S\otimes\nu_T(C)=0$, then from Fubini's theorem $$0=\int_S\nu_T(C_s)\,\nu_S(ds)=\int_T\nu_S(C^t)\,\nu_T(dt)$$ where $C_s=\{t\in T: (s, t)\in C\}$ and $C^t=\{s\in S: (s, t)\in C\}$. Hence, $\nu_S$-almost all $s\in S$, $\nu_T(C_s)=0$ (similarly, for $\mu_T$-almost all $t\in T$, $\nu_S(C^t)=0$). Consequently, for $\nu_S$-almost all $s\in S$, $\int_{C_s}f(s, t)\,\nu_T(dt)=0$ $$\nu(C)=\int_C f\,d\nu_T\otimes d\nu_S=\int_S\Big(\int_{C_s}f(s, t)\nu_T(dt)\Big)\,\nu_S(ds)=0$$ Therefore, under assumptions (1) ans (2) $\nu\ll\nu_S\otimes\nu_T$