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Suppose $\{f_n\}$ are Lebesgue measurable functions on $[0,1]$, such that $\int_0^1 |f_n|\,d\mu=1$ for all $n$, and $f_n\to 0$ almost everywhere.

I have proved: given $\epsilon>0$, there exists a Lebesgue meausurable $E\subseteq [0,1]$ such that $\mu(E)<\epsilon$ and $$\lim_{n\to\infty}\int_E |f_n|\,d\mu=1$$ (using Egorov's Theorem, where $E$ turns out to be $[0,1]\setminus F$ for some closed $F$ in which the convergence is uniform.)

Hence, or otherwise, how do we prove that there exists a subsequence $f_{n_k}$ of $f_n$ and sequences of measurable functions $g_k$ and $h_k$ such that

(i) $f_{n_k}=g_k+h_k$ for all $k$

(ii) $g_kg_j=0$ a.e. for $k\neq j$

(iii) $\lim_{k\to\infty}\int_0^1|h_k|\,d\mu=0$


The hardest condition in my opinion is (ii). I managed to find a candidate that satisfies both (i) and (iii), but not (ii). Let $f_{n_k}$ be a subsequence such that $\int_E |f_{n_k}|\,d\mu>1-\frac 1k$. Let $g_k=f_{n_k}\chi_E$ and $h_k=f_{n_k}\chi_F$, where $E=[0,1]\setminus F$.

Then (i) $f_{n_k}=g_k+h_k$ is satisfied.

$\lim_{k\to\infty}\int_0^1 |h_k|\,d\mu=\lim_{k\to\infty}\int_F |f_{n_k}|\,d\mu=1-\lim_{k\to\infty}\int_E |f_{n_k}|\,d\mu=1-1=0$. So (iii) is satisfied.

However, condition (ii) remains unsatisfied.

Thanks for any help.

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    $\begingroup$ You should also find a sequence of sets $E_{k+1}\subset E_k$, such that $\mu(E_k)>0$ and $\mu(E_{k+1})/\mu(E_k)\to 0$ as $k\to\infty$. Applying your idea to $E_{k}\backslash E_{k+1}$ will work I think $\endgroup$ – Del Oct 28 '16 at 10:07
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Using Egorov, we can choose $E_1\subset [0,1]$ such that $\int_{E_1}|f_1| > 0$ and $f_n\to 0$ uniformly on $E_1.$ Now

$$1= \int_0^1|f_n| = \int_{E_1}|f_n| + \int_{[0,1]\setminus E_1}|f_n|,$$

and since $f_n\to 0$ uniformly on $E_1,$ we have $\int_{[0,1]\setminus E_1}|f_n|\to 1.$ Thus there exists $n_2>1$ such that $\int_{[0,1]\setminus E_1}|f_{n_2}| > 1/2.$ By Egorov we can choose a sequence $F_k \subset [0,1]\setminus E_1$ such that

$$\mu (F_k) \to \mu([0,1]\setminus E_1),$$

with $f_n \to 0$ uniformly on each $F_k.$ So if $k$ is large enough, we'll have both $f_n \to 0$ uniformly on $F_k$ and

$$\int_{F_k}|f_{n_2}| > 1/2.$$

Let $E_{2}$ be be any one of these $F_k.$ So we now have pairwise disjoint $E_1,E_2$ and $1= n_1 < n_2$ such that $f_n \to 0$ uniformly on $E_1\cup E_2$ and $\int_{E_k}|f_{n_k}| > 1-1/k$ for $k=1,2.$

We can continue this process by induction to obtain pairwise disjoint subsets $E_1, E_2, \dots$ and $1=n_1 < n_2 < \cdots $ such that for each $k,$ $f_n \to 0$ uniformly on $E_k$ and $\int_{E_k}|f_{n_k}| > 1-1/k.$

Now it's easy street. Define $g_k = f_{n_k}\chi_{E_k}, h_k = f_{n_k}\chi_{[0,1]\setminus E_k}.$ Then (i)-(iii) are satisfied.

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  • $\begingroup$ Thanks. I understand most of it. However, in the first line, how can we guarantee that there is $\int_{E_1}|f_1|>0$? $\endgroup$ – yoyostein Nov 6 '16 at 16:52
  • $\begingroup$ I was hoping someone would ask that; I left it out on purpose. It's exactly what we do with the $F_k$'s later on. We know $\int_0^1 |f_1| = 1.$ We also know, by Egorov, that there exist $F_k \subset [0,1]$ such that $\mu(F_k) > 1-1/k $ and $f_n \to 0$ on each $F_k.$ Verify that $\int_{F_k} |f_1| \to 1.$ Thus any $F_k$ with that integral $>0$ will do for $E_1.$ $\endgroup$ – zhw. Nov 6 '16 at 17:30

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