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Attention

The usual problems are about absolute convergence: $$\int|g_n|\mathrm{d}\mu\quad(g_n=f_n,f-f_n,s_m-s_n,\ldots)$$

(There Fatou may help out!)

But as proceeding with Fatou one encounters that one can't distort to the limessuperior: $$\int\limsup_n|g_n|\mathrm{d}\mu\nleq\int\liminf_n|g_n|\mathrm{d}\mu\leq\liminf_n\int|g_n|\mathrm{d}\mu\leq\limsup\int|g_n|\mathrm{d}\mu$$

So the real question is about the analogue for limessuperior!!

(And not the analogue for negative functions...)

Problem

Given a measure space $\Omega$.

The lemma of Fatou states: $$f_n\geq0:\quad\int\liminf_nf_n\mathrm{d}\mu\leq\liminf_n\int f_n\mathrm{d}\mu$$ Does the reverse hold true: $$f_n\geq0:\quad\int\limsup_nf_n\mathrm{d}\mu\leq\limsup_n\int f_n\mathrm{d}\mu$$ Certainly, for convergent examples this holds true: $$f_n\geq0:\quad\int\limsup_nf_n\mathrm{d}\mu=\int\lim_nf_n\mathrm{d}\mu\leq\lim_n\int f_n\mathrm{d}\mu=\limsup_n\int f_n\mathrm{d}\mu$$ So one needs to dig deeper to find an honest counterexample!!!!

(I intend to answer my own question!)

(I have to admit that my earlier answer was lame!)

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    $\begingroup$ what is the point of asking a question and posting the answer right afterwards ? $\endgroup$ – mookid Dec 12 '14 at 22:54
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    $\begingroup$ @mookid Can I answer my own question? Yes! Stack Exchange has always explicitly encouraged users to answer their own questions. If you have a question that you already know the answer to, and you would like to document that knowledge in public so that others (including yourself) can find it later, it's perfectly okay to ask and answer your own question on a Stack Exchange site. $\endgroup$ – user147263 Dec 12 '14 at 22:57
  • $\begingroup$ @mookid: The point is to remind myself (and others). But please have a look on: Answer own Question $\endgroup$ – C-Star-W-Star Dec 12 '14 at 22:58
  • $\begingroup$ @mookid: You're welcome. ;) Unfortunately, many people don't ask about it before judging. :( $\endgroup$ – C-Star-W-Star Dec 12 '14 at 23:16
  • $\begingroup$ about your question: this is the whole point of the hypothesis $f\ge 0$ in the original Fatou lemma. $\endgroup$ – mookid Dec 12 '14 at 23:28
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Attention

The problematic examples are the divergent ones: $\liminf\neq\limsup$

Counterexamples

On the one hand, it holds by the usual Fatou: $$f_n:=\chi_{(n,n+1]}:\quad0=\int\limsup_nf_n\mathrm{d}\mu\leq\limsup_n\int f_n\mathrm{d}\mu=1$$ On the other hand, it fails the reverse Fatou: $$f_{1\leq k\leq n}:=\chi_{(\frac{k}{n},\frac{k}{n}]}:\quad1=\int\limsup_nf_n\mathrm{d}\mu\nleq\limsup_n\int f_n\mathrm{d}\mu=0$$ (So while the former forces the direction of the inequality the latter shows it can't hold.)

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  • $\begingroup$ I don't understand your indexing of $f_n$ in $$f_{1\leq k\leq n}:=\chi_{(\frac{k}{n},\frac{k}{n}]}$$ $\endgroup$ – Duchamp Gérard H. E. Dec 29 '16 at 21:38
  • $\begingroup$ The $\{1\leq k\leq n\}:=\{(k,n):1\leq k\leq n\}$ run as $(k,n)=(1,1),(1,2),(2,2),\ldots,(1,5),\ldots(5,5),(1,6),\ldots,(6,6),\ldots$ $\endgroup$ – C-Star-W-Star Dec 30 '16 at 10:44
  • $\begingroup$ So, in my language, it is $$g_n=\sum_{1\leq k\leq n}\chi_{[\dfrac{k}{n},\dfrac{k}{n}]}$$. Is what you want to define ? $\endgroup$ – Duchamp Gérard H. E. Dec 31 '16 at 6:13
  • $\begingroup$ No, the indexing here does not consist of only a single "number" $\lambda:=n$ but really of tuples of "numbers" $\lambda:=(nk)$. You can think of it is a diagonally counting the tuples similar to the proof that the rationals are countable: $1'=(11), 2'=(12), 3'=(22), 4'=(13),\ldots$. $\endgroup$ – C-Star-W-Star Dec 31 '16 at 15:04
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    $\begingroup$ In general, as you will hopefully learn later on, one is not limited to only take the natural numbers as indexing set (sequences) but any (directed) preordered set (net) will do a job too. This veeery often turns technical matters into natural ones. Also it comes in very handy in many constructions and applications as for example defining arbitrary sums/integrals. $\endgroup$ – C-Star-W-Star Dec 31 '16 at 15:10

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