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Let $a_n=\sum_{k=1}^n \dfrac{1}{k}$, $a_0=0$. We define $$f_n:\mathbb R \to \mathbb R$$$$f_n(x)=\sum_{k=1}^n \dfrac{1}{2^{n-k}} \mathcal X_{[a_{k-1},a_k)}(x)$$

Decide if there exists $f$ such that $f_n \stackrel{m}{\to}f$

I am not sure what to do here. It is clear that this sequence is increasing, if there is a function $f$ such that the sequence converges to $f$ in measure, then I could extract a subsequence that converges pointwise almost everywhere to $f$. Then I could apply the monotone convergence theorem to affirm that the limit of the Lebesgue integral of the sequence is equal to the integral of $f$.

I don't know what to do and if I could arrive to a contradiction following the path I've previously described. Any help would be appreciated. Thanks in advance.

saz
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user156441
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1 Answers1

3

Hints:

  1. Since $f_n(x)=0$ for any $x<0$, we have $f_n(x) \to 0$ as $n \to \infty$ for $x<0$.
  2. For any $x \geq 0$, there exists a unique $k=k(x) \in \mathbb{N}$ such that $x \in [a_{k-1},a_k)$. Show that $$f_n(x) = \frac{1}{2^{n-k}}$$ for all $n \geq k$.
  3. Conclude that $f_n(x) \to 0$ as $n \to \infty$ for any $x \geq 0$.
saz
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  • Thanks for the help. In 2., I don't see why for ANY $x \geq 0$ this works, but I think it is not necessary to show that for all $x \geq 0$ this happens since if $x$ is not in any of the intervals, then $f_n(x)=0$ for all $n$. I've said that $f_n$ is increasing which is wrong, but as you've said, $f_n(x) \to 0$ when $n \to \infty$. If I could show that $\int \lim f_n=\lim \int f_n$, then $0=\lim \int f_n$. So we would have $m({x: f_n(x)>\delta})\leq \dfrac{1}{\delta}\int f_n \to 0$ when $n \to \infty$, so $f_n \stackrel{m}{\to}f$. – user156441 Sep 13 '15 at 01:02
  • @user156441 Note that $(a_n){n \in \mathbb{N}}$ is a strictly increasing sequence satisfying $a_n \to \infty$ as $n \to \infty$. Therefore, there exists for any $x \geq 0$ some $k$ such that $x \in [a{k-1},a_k)$. Concerning your other remark: Yes, $L^1$-convergence implies convence in measure, but it is not necessary (i.e. there exist sequences which converge in measure to $0$, but not in $L^1$; see http://math.stackexchange.com/q/1170559/) – saz Sep 13 '15 at 05:53