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| visits | member for | 2 years, 1 month |
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1d |
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space of riemann integrable functions not complete @Tim Then I should find other examples. But then the above example is just saying that there is a (cauchy) sequence in $\mathcal{R}^1$ such that the pointwise limit is not in $\mathcal{R}^1$? |
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May 5 |
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Is it necessary to consider inversion $\left(z\mapsto\dfrac{1}{z}\right)$ on the extended complex plane @EricStucky I think that the answer depends on what is the definition of a straight line is. So if the definition does not need to contain the infinity point, you don't need to consider the extended complex plane. |
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Apr 23 |
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integral on $[a,x]$ is zero for all x implies $f=0$ a.e. @jun Oh I couldn't find it. I should take a look at it. |
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Apr 23 |
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integral on $[a,x]$ is zero for all x implies $f=0$ a.e. @Easy yes, I'll fix it. |
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Apr 22 |
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A function with countable discontinuities is Borel measurable. Actually I didn't proved the last statement, so there can be some errors. |
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Apr 22 |
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A function with countable discontinuities is Borel measurable. @DaveL.Renfro The solution start in there. $f$ is then riemann integrable, so I can define the partition to be increasing set of discontinuous points $P_k$ and lower and upper (Borel-)simple functions $g_{P_k},G_{P_k}$ and the limits of them $g,G$. And use the fact that for $x\in \cup P_k$, $f$ is continuous at $x$ iff $g(x)=G(x)$. |
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Apr 22 |
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A function with countable discontinuities is Borel measurable. The generalization should be that if all subsets of {discontinuies} are measurable. |
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Apr 22 |
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A function with countable discontinuities is Borel measurable. Or maybe the conclusion is that if $f$ is satisfying such conditions, then $E$ should be measurable? |
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Apr 22 |
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A function with countable discontinuities is Borel measurable. I have a question. Your answer can be generallized to for any $f:E \to \mathbb{R}$ with the set of discontinuities measurable, then $f$ is measurable. But suppose that $E$ is not measurable. Your proof does not use any condition about $E$. Then $E=\cup \{ f<n \}$ should be measurable. What is the problem in my reasoning? |
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Apr 22 |
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A function with countable discontinuities is Borel measurable. This is so great! Thank you. |
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Apr 19 |
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Transitive set ordered by epsilon Thanks for great help. |
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Apr 19 |
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Transitive set ordered by epsilon Thanks, I should study more about ordinals. |
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Apr 19 |
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Transitive set ordered by epsilon But can you give me any hints for (b), last question? I'm having trouble in them, too. |
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Apr 19 |
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Transitive set ordered by epsilon I got it. $w=z$ or $z \in w$ both lead to $z \in (x \cap y)\cap(y-x)=\emptyset$. |
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Apr 18 |
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Product sigma algebra of Borel sigma algebra and Power set. @StefanHansen I edited it. |
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Apr 18 |
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Product sigma algebra of Borel sigma algebra and Power set. @StefanHansen Well, the sigma finiteness gives the unique product measure. So it may not be uniquely defined. |
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Mar 16 |
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Finitely generated torsion module over a PID. There was a basis, dual basis to start with, and also used that in finite dimensional vector space, injectivity implies surjectivity. |
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Mar 10 |
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smallest sigma algebra can be obtained by countable operations? And I guess that David answered "no" to my comment. |
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Mar 10 |
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smallest sigma algebra can be obtained by countable operations? @ZevChonoles Actually I didn't explained it well. But I'm just thinking of it as follows. From $\mathcal{A}$, let $\mathcal{A}(1)$ be the set obtained by countable unions, countable intersections, complements of $\mathcal{A}$. Similarly, $\mathcal{A}(n+1)$ be defined from $\mathcal{A}(n)$. So we get $\mathcal{A}\subset \mathcal{A}(1) \subset \mathcal{A}(2) \subset \cdots$. Then is it true that $\sigma(\mathcal{A})$=$\mathcal{A}(N)$ for some $N$? Or $\sigma(\mathcal{A})=\cup \mathcal{A(n)}$? |
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Dec 16 |
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To prove divergence of a series. @Amr As r tends to 1, the sum tends to infinity. |
