# integration and convergence of sets

I am a bit stuck proving this. Anyone has an idea? or a place I should look at?

Let $X\subset\mathbb{R}$ and $Y\subset\mathbb{R}$ be compact sets. Let $f:X\times Y\rightarrow\mathbb{R}$ be a $C^{1}$ function. Let $s:Y\rightarrow X$ be a function (not necessarily continuous). Define $m:X\times\mathbb{R}\rightarrow\mathbb{R}$ as: $m(x,h)=\int_{S(x,h)}f(x+h,y)dy$

where $S(x,h)= [ y \in Y:x \leq s(y) < x+h ]$ with $h>0$ and small.

Finally, $\forall(x,y)\in X\times Y$ such that $s(y)=x,f(x,y)=0$.

Question: Calculate the limit as $h\rightarrow0$ of $m(x,h)$

Thanks everyone!

-

I assume that there is some $\delta>0$ such that $[x,x+\delta]\subseteq X$.
If $s$ is measurable, the answer is $0$. This is because $m(x,h)$ can be rewritten as $$\int_Y f(x+h,y) \chi_{s^{-1}([x, x+h))}(y) dy,$$ and by assumption, the integrand converges pointwise to $0$. (Here, $\chi_{s^{-1}([x, x+h))}$ is the indicator function of $s^{-1}([x,x+h))$.)
If $s$ is not measurable, the integral need not be defined.
David: thanks very much but I do not know whether $s^-1$ exists. nothing guarantee that the function is invertible. thanks a lot for the help! sandy – sandybrooks Jan 29 '13 at 23:30
For any function $s$ and subset $S$ of its range, the set $s^{-1}(S)$ is defined to be the set of all $x$ such that $s(x)\in S$. This does not assume that $s$ is invertible. – David Moews Jan 29 '13 at 23:54