Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question

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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.