I am trying to solve a qualifying exam problem.

I would like to know what can be said about the convergence of the indicator functions $I_{E_k}$ in $L^2[0, 1]$ when it's known that $m(\limsup E_k\setminus\liminf E_k) = 0$.

My guess is that at least $I_{E_k}$ converges in $L^2$, and it seems like one way to show it is to show that $I_{E_k}$ is Cauchy, using the definitions of the lim sup and lim inf of a sequence of sets, but this is where I'm stuck.

Any hints would be appreciated!

Edit: Basically I'd like to make this rigorous:

$$\|I_{E_k} - I_{E_l}\|^2_2 = m(E_k \Delta E_l) = m((E_k \cup E_l) \setminus (E_k \cap E_l)).$$

For $l > k$ and as $k\rightarrow\infty$,

$$ \begin{align*} 0 &\leq m((E_k \cup E_l) \setminus (E_k \cap E_l)) = \lim\sup_k \int I_{(E_k \cup E_l) \setminus (E_k \cap E_l)}dm \\ &\leq \int \lim\sup_k I_{(E_k \cup E_l) \setminus (E_k \cap E_l)}dm \leq \int I_{(\lim\sup E_n \setminus \lim\sup E_n)}dm \\ &= m(\lim\sup E_n \setminus \lim\inf E_n) = 0, \end{align*}$$ where we use Fatou's lemma when moving the lim sup inside the integral.

Since $\{I_{E_k}\}$ is Cauchy, it converges.

  • $\begingroup$ I would proceed differently: the condition you are given says that the sequence converges a.e. By DCT it converges in $L^1$. But for this kind of functions the $L^2$ distance is just the square root of $L^1$ distance. $\endgroup$ – user53153 Dec 21 '12 at 7:44
  • $\begingroup$ Pavel--thanks for the great suggestion. Do you think it's possible to make this work in a general finite measure space, where it is not guaranteed that the measure space is complete? $\endgroup$ – Javalina Dec 21 '12 at 18:48
  • $\begingroup$ I would apply the argument to the completion, obtaining $L^2$ convergence there. Then see what it means for the original incomplete space. $\endgroup$ – user53153 Dec 21 '12 at 19:24

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