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I need some help o understand the construction of measure induce by a function. For $d=1$, given right-continous function $F:\mathbb{R}\rightarrow [0,1]$ that increasingly monotonic such that $F(\infty)=1$ and $F(-\infty)=0$ we can defined a measure $\mu$ on Borel set by extend (using Caratheodory extension theorem) a function $\nu:\{(a,b]\}\rightarrow [0,1]$ defined by $\nu((a,b])=F(b)-F(a)$. For this case, I have proved that $\nu$ satisfies $\sigma$-additive properties and $\nu(\mathbb(R)<\infty$ so we can extend $\nu$ to become a unique measure $\mu$

I need help to prove for the case $d=2$. Given a continous function $F:\mathbb{R}^2 \rightarrow [0,\infty)$ that increasing ($F(a_1,b) \leq F(a_2,b)$ for $a_1\leq a_2$ and $F(a,b_1) \leq F(a,b_2)$ for $b_1\leq b_2$) and satisfy $F(b_1,b_2)-F(a_1,b_2)-F(b_1,a_2)+F(a_1,a_2)\geq 0$ for $a_1\leq b_1 $ and $a_2 \leq b_2$ and also $F(-\infty,-\infty)=0$ and $F(\infty,\infty)=1$. Could we get a unique measure $\mu$ on Borel $B(\mathbb{R}^2)$ induced by $F$ such that $\mu((x_1,y_1] \times (x_2,y_2])=F(y_1,y_2)-F(x_1,y_2)-F(y_1,x_2)-F(x_1,x_2)$. I know that we have to show that $\nu$ defined by $\nu((x_1,y_1] \times (x_2,y_2])=F(y_1,y_2)-F(x_1,y_2)-F(y_1,x_2)+F(x_1,x_2)$ must satisfies $\sigma$-additive properties and $\sigma (\mathbb{R}^2)=1$ but I dont know the detail.

Thanks.

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First of all I think $\nu$ is not well defined, because when $x_{1}\to y_{1}$ and $x_{2}\to y_{2}$ then $\nu<0$, maybe $\nu$ should be defined $$\nu((x_1,y_1] \times (x_2,y_2])=F(y_1,y_2)-F(x_1,y_2)-F(y_1,x_2)+F(x_1,x_2),$$ that according the hypothesis is positive, . Now this measure assigns to each "rectangle" the sum of the values of $F$ in the left-down and right-up corners minus the sum of the values in the left-up and right-down corners, then when you consider two rectangles $A$, $B$ having one side in common then $A\cup B$ is a rectangle and $\nu(A\cup B)=\nu(A)+\nu(B)$, maybe this can help you.

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