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Let E_n be a sequence of Lebesgue measureable sets in [0,1]. Suppose that for $0 \leq k \leq 1$ we have that $m(E_n \cap [0,r])= kr$ for any r, such that $0 \leq r \leq 1$.

Prove that the $$\lim_{ n \rightarrow \infty} \int_{E_n} f(x) dx= k \int_{[0,1]} f(x) dx,$$ where $f \in L^1([0,1])$.

I have attempted the following. $$k \int_{[0,1]} f(x) dx = \lim_{n \rightarrow \infty} \frac{m(E_n \cap [0,r])}{r} \int_{[0,1]} f(x)dx = \lim_{n \rightarrow \infty} \int_{[0,r]} \frac{\chi_{E_n} (t)}{r} dt \int_{[0,1]} f(x) dx= \int_{[0,1]} \int_{[0,r]}\frac{\chi_{E_n} (t) f(x)}{r} dt dx.$$

I want to somehow change the order of integration to change $\chi_{E_n}(t)$ to $\chi_{E_n}(x)$ (perhaps applying Fubini Tonelli), but I don't think its possible.

*******Applying the comments suggestions********

Since step functions are dense in $L^1$ there exist $\phi_l \nearrow f$. We note that we can apply DCT because $\phi_l \leq f$ and $f \in L^1$.

$$\lim_{n \rightarrow \infty} \int_{[0,1]} \chi_{E_n}(x) \lim_{l \rightarrow \infty} \phi_l(x)dx= \lim_{n \rightarrow \infty} \lim_{l \rightarrow \infty} \int_{[0,1]} \chi_{E_n}(x) \phi_l(x)dx.$$

I want the change the order of the limits to say that $$\lim_{n \rightarrow \infty} \lim_{l \rightarrow \infty} \int_{[0,1]} \chi_{E_n}(x) \phi_l(x)dx= \lim_{l \rightarrow \infty} \lim_{n \rightarrow \infty} \int_{[0,1]} \chi_{E_n}(x) \phi_l(x)dx= \lim_{l \rightarrow \infty} k \int_{[0,1]} \phi_l (x) dx = k \int_{[0,1]} f(x)dx.$$

I don't know how to justify that I can indeed change the order of the limits.

Mark Fantini
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Lucy Lewis
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  • Prove it in the special case that $f$ is a step function. Then show that it is sufficient to prove it for step functions. – Stephen Montgomery-Smith Jun 04 '14 at 03:20
  • I was able to show that it held for simple functions. To show that is it is sufficient to prove for step functions requires that I change two limits. I am not sure how justify that I can do that. To show you what I did I will edit the entry of the question because I am running out of characters. – Lucy Lewis Jun 04 '14 at 04:06
  • You want to start it like this: since the step functions are dense in $L^1$, given $\epsilon>0$, there exists a step function $\chi$ such that $\int|\chi(x)-f(x)|, dx < \epsilon$. – Stephen Montgomery-Smith Jun 04 '14 at 04:16
  • Note also that not all simple functions are step functions. By step function I mean that it is a finite linear combination of characteristic functions of intervals. – Stephen Montgomery-Smith Jun 04 '14 at 04:17
  • Yes, I know sorry about that. I think I got it work. Thank you! – Lucy Lewis Jun 04 '14 at 04:44

1 Answers1

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I do not see the importance of the index $n$ in this problem, you can show in fact that for every $n\in \mathbb{N}$, we have $$\int_{E_n} f(x) dx = k \int_0^1 f(x) dx,$$ which implies the equality will also hold if we take the limit as $n\rightarrow \infty$.

For now let us call $E_n$ simply $E$ since the equality does not depend on the index $n$.

Define a measure $\mu_E:\mathcal{A} \rightarrow \mathbb{R}$ such that $$\mu_E (A) = m(A\cap E).$$ (check this is indeed a measure)

From the assumption that $m(E\cap [0,r]) = kr,$ we can apply Dynkin $\pi - \lambda$ theorem to conclude that $$\mu_E (A) = m(A\cap E) = km(A)$$ for each $A\in \mathcal{A}$. Then the integral equality would hold trivially.

Edit: $m(E\cap [0,r]) = kr,$ implies that $$\mu_E (A) = m(A\cap E) = km(A)$$ for each $A\in \mathcal{A_0}$, the collection of all finite union of intervals in $[0,1]$. If you are not familiar with the Dynkin $\pi - \lambda$ theorem, see here (I used it to show that two measures are the identical if they agree on each interval).

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