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I have the following problem: Show $\beta \ll \eta$ if and only if for every $\epsilon > 0 $ there exists a $\delta>0$ such that $\eta(E)<\delta$ implies $\beta(E)<\epsilon$.

For the forward direction I had a proof, but it relied on the use of the false statement that "$h$ integrable implies that $h$ is bounded except on a set of measure zero".

I had no problem with the backward direction.

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    Then what exactly is the question? $\beta << \eta \Rightarrow \varepsilon,\delta$ statement? And the definition of absolutely continuos is $\beta = f\eta$ with $\eta$-integrable $f$? If yes, use characteristic functions of measurable sets.2012-08-19
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    The question is how to do the forward direction, so yes.2012-08-19
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    And what exactly do you mean with the characteristic functions?2012-08-19
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    To do the forward direction you need some restriction on $\beta$. The following example is from Rudin's "Real & Complex Analysis": Take $X=(0,1)$, $\eta$ the Lebesgue measure and $\beta(E) = \int_E \frac{1}{t} dt$. Then $\beta \ll \eta$, but the $\epsilon-\delta$ characterization doesn't hold.2012-08-19
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    What if our set is of finite measure?2012-08-19
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    That will work. There was an answer here that was correct except for a minor point: If $E_{n+1} \subset E_n$, then to conclude that $\lim \mu E_n = \mu \cap E_n$, at least one of the sets must have finite measure.2012-08-19
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    Meaning finite measure with respect to $\beta$. And Rudin's example was after Theorem 6.11 (3rd Ed.).2012-08-19
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    @copper.hat: thanks for that example. I noticed that I implicitly assumed $\beta$ to be totally finite and tried to find a fix for $\sigma$-finite $\beta$, but it looked like a complete mess, no wonder...2012-08-19
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    @t.b.: Rudin deserves the credit! I used to have a nice book called "Counterexamples in Analysis" by Gelbaum & Olmsted, but I haven't been able to locate it for years now.2012-08-19
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    The statement of the problem should begin by saying what $\beta$ and $\eta$ are.2012-08-19

2 Answers 2

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Assume that $\beta=h\eta$ with $h\geqslant0$ integrable with respect to $\eta$, in particular $\beta$ is a finite measure. Let $\varepsilon\gt0$.

There exists some finite $t_\varepsilon$ such that $\beta(B_\varepsilon)=\int_{B_\varepsilon} h\,\mathrm d\eta\leqslant\varepsilon$ where $B_\varepsilon=[h\geqslant t_\varepsilon]$. Note that, for every measurable $A$, $A\subset B_\varepsilon\cup(A\setminus B_\varepsilon)$, hence $\beta(A)\leqslant\beta(B_\varepsilon)+\beta(A\cap[h\leqslant t_\varepsilon])\leqslant\varepsilon+t_\varepsilon\eta(A)$.

Let $\delta=\varepsilon/t_\varepsilon$. One sees that, for every measurable $A$, if $\eta(A)\leqslant\delta$, then $\beta(A)\leqslant2\varepsilon$, QED.

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    For the line where "$\beta(A) \leq \beta(h \geq t_{\epsilon})...$, could you explain this a bit more? I'm having problems seeing where that comes from.2012-08-19
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    So, for any measurable set $A$, and given an $\epsilon>0$ we have that $\beta(A)=\int_Ahd\eta$ integrable implies that we can find a $t_{\epsilon}$ such that $\beta(A_{\epsilon})\leq \frac{\epsilon}{2}$, with $A_{\epsilon}=\lbrace x\in A| h(x) \geq t_{\epsilon} \rbrace$. Then $A_{\epsilon}^C=\lbrace x\in A| h(x) < t_{\epsilon} \rbrace$. So, taking $\delta<\frac{\epsilon}{2t_\epsilon}$ we have that $\beta(A_{\epsilon})\leq \beta(A_{\epsilon}) + \beta(A_{\epsilon}^C) \leq \frac{\epsilon}{2} + t_{\epsilon}\eta(A_{\epsilon}^C$).2012-08-19
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    Now, if $\eta(A_{\epsilon})< \delta$ then we have from above that $\eta(A_{\epsilon}^C)\leq \eta(A_{\epsilon})< \delta$, thus $\beta(A_{\epsilon}) \leq \frac{\epsilon}{2}+t_{\epsilon}\frac{\epsilon}{2t_{\epsilon}}=\epsilon$. Is this correct?2012-08-19
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    Note that this answer assumes that $\beta$ is finite. This is a reasonable assumption, but it is not immediately obvious.2012-08-19
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    @copper.hat Indeed, requiring $h$ to be integrable is assuming that $\beta$ is finite and I could have made this more obvious (thanks).2012-08-19
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    @FrankWhite See edited version.2012-08-19
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    @did, thanks for the proof. Just a small question: from $\beta \ll \eta$, how can we make the assumption $\beta = h\eta$, for integrable $h$? Thanks again!2012-10-23
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    @RVC This is one chararacterization of the absolute continuity.2012-10-23
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No need to use Radon-Nikodým here. I'll assume that $\beta$ is totally finite. See the end of the answer why this is necessary.

Suppose that $\beta E = 0$ whenever $E$ is measurable and $\eta E = 0$ but that the desired $\varepsilon$-$\delta$-condition doesn't hold.

Then there is $\varepsilon \gt 0$ such that for all $\delta \gt 0$ there is $E$ such that $\eta(E) \lt \delta$ but $\beta(E) \geq \varepsilon$. For each $n$ choose $E_n$ such that $\eta(E_n) \lt 2^{-n}$ and $\beta(E_n) \geq \varepsilon$.

Define $E = \bigcap_{N \in \mathbb{N}} \bigcup_{n \geq N} E_n$. Then $$ 0 \leq \eta E \leq \inf_{N \in \mathbb N} \eta \bigcup_{n \geq N} E_n \leq \inf_{N \in \mathbb{N}} \sum_{n \geq N} 2^{-n} = 0 $$ hence $\eta E = 0$. By hypothesis it follows that $\beta E = 0$ as well. On the other hand, assuming $\beta$ is totally finite, we get $$ 0 = \beta E = \lim_{N \to \infty} \beta \bigcup_{n \geq N} E_n \geq \varepsilon \gt 0, $$ which is absurd.


Note that, as @copper.hat pointed out in the comments, it is necessary to assume that $\beta$ is totally finite. The $\sigma$-finite example $\beta E = \int_{E} \frac{1}{t}\,dt$ on $(0,1)$ shows this: For Lebesgue measure $\lambda$ on $(0,1)$, the absolute continuity condition “$\lambda E = 0$ implies that $\beta E = 0$” holds, while the $\varepsilon$-$\delta$-characterization doesn't. For every $\delta \gt 0$ we have $\beta(0,\delta) = \infty$ while $\lambda(0,\delta) = \delta$.

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    That $\eta(E) = 0$ follows also from Borel-Cantelli lemma. (actually, one can easily prove this lemma using your argument)2012-08-19
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    That's right, but I tried to keep this argument as elementary as possible. Thanks!2012-08-19