This relates to a question I had asked here before.

Suppose we have a density $\phi$ on $\mathbb{R}^{2}$. Let $\Phi$ be the Borel associated measure (is this the right term?). That is,

\begin{equation} \int_{A\times B} \phi(x,e)dxde = \int_{A\times B} d\Phi = \Phi(A\times B) \end{equation}

Now here is my question. Is there a Borel measure $\Lambda$ associated(?) with the function $x\phi(x,e)$ such that

\begin{equation} \Lambda(h) = \int hd\Lambda = \int h(x,e) x\phi(x,e)dxde \end{equation}

where $h$ is a measurable function on $\mathbb{R}^{2}$. My intuition tells me that $\Lambda$ is a measure satisfying:

\begin{equation} \int_{B\times A}d\Lambda = \int\mathbb{1}_{B\times A}{(x,e)}x\phi(dx,de)dxde \end{equation}

Is this correct?

This is my attempt at proving the integration:

\begin{equation} \int d\Lambda = \int h(x,e)\int^{dx}\int^{de}a\phi(a,y)dady = \int h(x,e) x\phi(x,e)dxde \end{equation}

where the second equality uses the fundamental theorem of calculus. Of course, I am not very confident about what I have done here.


1 Answer 1


Note first that you cannot expect a positive measure. You get a signed measure provided $x\phi(x,e)$ is integrable on $\mathbb R^{2}$. To prove this you define $\Lambda (E)$ as $\int I_E(x,e)x\phi(x,e)dxde$ for any Borel set $E$ in $\mathbb R^{2}$ (not necessarily measurable rectangles) and $\Lambda$ will be a signed measure. If $\phi (x,e)=0$ for $x<0$ then $\Lambda$ will be a positive finite measure.

  • $\begingroup$ Ok. but then how do I integrate wrt to $\Lambda$? Let us assume $\phi(x,e) = 0$ for $x<0$. $\endgroup$ Apr 2, 2018 at 10:13
  • $\begingroup$ Just apply definition of $\Lambda$ to prove the formula for $\int h d \Lambda$ for a simple function $h$. It then holds for non-negative measurable functions (as well as integrable functions) . $\endgroup$ Apr 3, 2018 at 8:53

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