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This is question 10.2 in "Measures, Integrals and Martingales". The answer, which can be found here seems to contain a typo, and also I don't see why they pick such a complicated counterexample. I would like to know if the approach taken in the solutions would work, and also if my counterexample is sufficient.

The question: Let $(\Omega,\mathcal{A},P)$ be a propability space. Find a counterexample to the claim: every P-integrable function $u \in \mathcal{L}^1$ is bounded. [Hint: you could try to take $\Omega = (0,1)$, $P = \lambda^1$, and show that $u = 1/\sqrt{x}$ is Lebesgue integrable on (0,1) by finding a sequence of suitable simple functions that is above u on, say, $(1/m,1)$, and then let $m \to \infty$ using Beppo Levi]

Now in the solutions, they use $u(\frac{j+1}{n}) = \sqrt\frac{j+1}{n}$, which should be the inverse and thus the following convergence argument doesn't apply.

But apart from that, why not just take an atom A (set with no subsets in the $\sigma$-algebra), construct a measure $\mu$ such that $\mu(A) = 1$, $\mu(A^c) = 0$ and set $u$ to 1 on $A$ and to infinity on $A^c$? That would be sufficient to falsify the claim right?

KCd
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Scipio
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  • Yes, if $u(x)=1/\sqrt{x}$ then $u((j+1)/n)=\sqrt{n/(j+1)}$. Does this rather typo deserve a question? Re why not the example you suggest, please note that both yours and the text example are interesting. Pointing out that the problem is not only with measures having atoms is necessary, in my opinion. – Did Nov 04 '15 at 12:36
  • Thank you! the typo on itself is not so interesting, but because of the typo I cannot check if their answer is correct - and also if my answer is lacking. Which is why I posted a question – Scipio Nov 04 '15 at 12:40
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    If your question is how to complete Exercise 10.2's solution, note that one is left with $$\frac1{\sqrt{n}}\sum_{j=1}^{n-1}\frac1{\sqrt{j+1}}\leqslant\frac1{\sqrt{n}} \int_1^{n}\frac{dx}{\sqrt{x}}=\frac1{\sqrt{n}}\left.2\sqrt{x}\right|_1^n<2.$$ – Did Nov 04 '15 at 12:40
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    ...or see this question or this question for an elementary proof. (... and I'll fix the typo.) – saz Nov 04 '15 at 13:17

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