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Expanding on @MichaelBurr's comment: $L^2[0,1]$ does not consist of functions, but rather of equivalent classes of functions. We say that two functions are in the same equivalence class if the set they disagree on is null. As such, when we say $1/\sqrt[4]{x}$ is a member of $L^2[0,1]$, we are saying it is in an equivalence class that contains a function ...

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\begin{align*} (a,\infty)=&\,\{x\in\mathbb R\,|\,x>a\},\\ (a,\infty]=&\,\{x\in\mathbb R\,|\,x>a\}\cup\{\infty\}. \end{align*} The latter set includes an extra point termed “positive infinity.” Note that it is not a real number, but in certain areas of mathematics, especially in measure theory, it is useful to extend $\mathbb R$ by this single ...

3

I still don't understand the question. Yes, if $f\ge0$ then all three integrals could be infinite. I don't understand why you feel that's a problem; Tonelli does not say the three integrals are finite, just that they are equal. Not quite understanding what the issue is, I'll try to explain the facts. NOTE we're assuming that $f$ is jointly measurable in ...

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Suppose that $S \cup N$ is measurable. Since $N$ is measurable, it follows that $N^c$ is measurable. Therefore $(S \cup N) \cap N^c = S \backslash N$ is measurable. Also, since $N$ has measure zero and since $S \cap N \subset N$, it follows by completeness of Lebesgue measure that $S \cap N$ is measurable. Therefore $S = (S \cap N) \cup (S \backslash N)$ ...

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Counterexample. In one dimension, let $K_k=\{0,1/k,2/k,3/k,\dots,1\}$. Each $K_k$ has zero measure. The limit of $K_k$ in the Hausdorff metric is the interval $[0,1]$, of measure $1$. You can also attach the same interval to the sets $K_k$ to make their measure positive. The above counterexample could be ruled out by requiring $K_k = ... 3 By the axiom of choice, there is a non-measurable set$A$contained in$[a,b]$. Then let$f=\chi_A$, the function defined as$\chi_A(x)=1$if$x \in A$and$\chi_A(x)=0$if$x \not\in A$. One can check that$f$is measurable if and only if$A$is measurable, which it's not. Then$f$cannot be Lebesgue integrable even though$\|f\|_{\infty}=1$. 3 I think I can confirm your suspicion that this doesn't necessarily hold if the target space is non-Hausdorff, assuming I haven't made a mistake somewhere... Let$\mathbb{R}$be the real line in its standard topology. Let$\mathbb{R}_0$be the real line with the topology whose non-empty open sets$U$are precisely the standard open sets$U \subseteq ...

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That "Wikipedia" definition is for the special case where the covariance matrix is $\sigma^2 I$. In that case they are equivalent. A measure $\mu$ on $\mathbb R^n$ has density $\rho$ with respect to Lebesgue measure $\lambda^n$ iff $\mu(A) = \int_A \rho(x) \; d\lambda^n(x)$ for all Lebesgue measurable sets $A$.

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We have $$f(X)\subseteq\{0\}\cup f(\mathrm{supp}(f)),$$ which is compact in $\mathbb{R}$ (since $\mathrm{supp}(f)$ is compact and $f$ is continuous), hence bounded.

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Every claim you started with is wrong. Your $S_{1}, S_{2}, S_{3}$ are all divergent series. It looks like you declared these series to be equal to their Ramanujan summation assignment, (which is akin to setting a matrix equal to its own determinant), and then proceeded as if you actually meant it converged to that value.

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Let $X=[0,1]$. Consider $1-f$. If $f\ne 1$ a.e. then $1-f\ne 0$ a.e. and indeed there is a positive measure set, $E$ so that $1-f>0$ on $E$. But then $$0=1-1=\int_X 1-\int_X f=\int_X(1-f)=\int_E (1-f)>0.$$

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For any sequence $f_n \to f''$ in $L^p$, there exists a subsequence such that $f_{n(k)} \to f''$ almost everywhere as $k \to \infty$. There are (at least) two possibilites to prove this statement: In the proof of the Riesz-Fischer theorem (which states that $L^p$ is a complete space), one usually constructs such a sequence (see e.g. René Schilling: ...

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Well, if the book doesn't include any finiteness hypotheses at all (look carefully!) then the book is simply wrong. If $\mu^*(A)=\mu^*(B)=\infty$ then $\mu^*(A)-\mu^*(B)$ is undefined. It's true if $\mu^*(A)=\infty$ and $\mu^*(B)<\infty$, by the way. In that case $\mu^*(A)\le\mu^*(B)+\mu^*(A\Delta B)$ shows that $\mu^*(A\Delta B)=\infty$. (It sounds ...

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The spectrum of the $C^*$-algebra $C_b(\mathbb R)$ of bounded continuous functions on $\mathbb R$ corresponds to the Stone-Čech compactification $\beta \mathbb R$ of $\mathbb R$. Any finite positive Borel measure on $\beta \mathbb R$ gives you a positive linear functional on $C_b(\mathbb R)$. However, the points of $\beta \mathbb R \backslash \mathbb R$ ...

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A lot of times these two mean the same thing, but it is important to consider the superset of which this is an interval. Sometimes, (especially in measure theory, which is why I mention it) it is useful to work in the extended reals, which includes a point at $\infty$, so $(a,\infty)$ means every number greater than $a$ accept infinity and $(a,\infty]$ ...

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I don't understand why nobody has upvoted Chilango's answer. His answer is correct except an obvious typo and that he supposed $E$ is a normed vector space in which case most applications of measurable functions occur. His proof is also correct when $E$ is a metric space if we replace the norm $|*|$ with a metric $d$. The following proof is just a detailed ...

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Let $\{A_n\}$ be a countable collection of sets and let $B$ denote the set of elements which are in $A_n$ for infinitely many $n$. Then $$B = \bigcap_{n=1}^{\infty}\bigcup_{k = n}^{\infty}A_k.$$ To see this, note that $x \in \bigcap\limits_{n=1}^{\infty}\bigcup\limits_{k = n}^{\infty}A_k$ if and only if $x \in \bigcup\limits_{k=n}^{\infty}A_k$ for all ...

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Let n be a large even number. Take n equally spaced points along the unit interval. A probability distribution which gives equal weight to all these points converges weakly to the uniform distribution. The product of two copies of this distribution converges weakly to uniform on the unit square. Since this distribution is equal to the product of it's ...

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Assuming a correct version of the standard definition of an outer measure $\mu^*$ given a measure $\mu$, and assuming that we're supposed to simply assuume that $\Omega_0\subset\Omega$ satisfies $\mu^*(\Omega_0)=\mu(\Omega)$ (that word "let" threw me off): Suppose $A\in F$. If we let $A_1=A$ and $A_n=\emptyset$ for $n>1$ then $A_n\in F$ and ...

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Actually no, a function has at most one Jordan decomposition. You have two decompositions, but at most one is the Jordan decomposition. Anyway, finiteness for one decomposition does not imply finiteness for the other. Consider $$0=0-0=x^+-x^+.$$ If $f$ has bounded variation and all your measures happen to be finite then yes, $\phi_1=\phi_2$. Because for ...

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The paper states directly before equation $2$ that $x_1, x_2, \cdots, x_n$ are uniformly drawn from the set $S$. The set $S$ was previously defined to be $B_{p, r}$, the $L^p$ ball of radius $r$ you mentioned above. The density of such as uniform random variable can be stated as: $$P(x_1, \cdots, x_n | p, r) = \frac{1}{\lambda(B_{p, r})}1_{(x_i \in ... 1 Assuming that Var(\epsilon_1)=\sigma_{\epsilon}^2 we have$$\mathbb{E}\beta^{2t}b_t^2= \sigma_{\epsilon}^2\frac{\beta^2}{(\beta^2-1)^2}+o(1)$$Hence, \beta^tb_t is L^2-bounded and uniformly integrable so that you can pass the limit inside the expectation... Edit: In the first step, you need to show that \beta^tb_t converges a.s. (to 0). This ... 1 Let A=\{(x,y)\in\mathbb{R}^2:x=y\}. You could use that B\otimes B are the Borel sets on \mathbb{R}^2 so a closed subset of \mathbb{R}^2 is measurable. If you don't know that yet you can go from first principles: Divide A into a countable union of A_n = \{(x,y)\in A: -n\leq x\leq n\}. Show that A_n is measurable by starting with the square ... 1 A measure on a product space need not be a product measure. Suppose$$ X = \left. \begin{cases} 0 & \text{with probability }1/4, \\ 1 & \text{with probability }3/4, \end{cases} \right\}\text{ and }Y = \left. \begin{cases} 0 & \text{with probability }1/3, \\ 1 & \text{with probability }2/3. \end{cases} \right\} \tag 1 $$The following two ... 1 Define$$f(t)=\sum_{n=-\infty}^\infty \hat\mu(n)e^{int}.$$The series converges uniformly so f is continuous. The uniform convergence also shows that$$\hat f(n)=\hat\mu(n).So uniqueness (for complex measures) shows that $\mu=f$, or more carefully $d\mu=f\,dt$. If you don't buy the uniqueness for complex measures bit: Lemma: If $\nu$ is a complex ...

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Brian M. Scott wrote: In the topological case I'd simply call it a (pairwise) disjoint family whose union is dense in $X$; I've not seen any special term for it. In fact, I can remember seeing it only once: such a family figures in the proof that almost countable paracompactness, a property once studied at some length by M.K. Singal, is a property ...

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If you do know the Weierstraß theorem, then you can prove it like that: Let $f \in \mathcal{K}(X)$ and denote by $K$ the support of $f$. Then $f|_{K^c}=0$ by the very definition of the support. Moreover, by the Weierstraß theorem, $f|_K$ is bounded. Combining both facts, proves that $f$ is bounded. If you do not know the Weierstraß theorem, then have e.g. ...

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A probability density function exists if and only if corresponding cumulative distribution function is absolutely continuous. A probability measure (or simply probability) is not a probability density function, but a measure (if a cumulative distribution function is specified, it is the corresponding Lebesgue-Stieltjes measure). Any random variable's ...

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It depends on how one defines $L^2(X)$. One way of defining $L^2(X)$ is as the set of all measurable functions $f:X\rightarrow\mathbb{R}$ such that $|f|^2$ is integrable over $X$. It is, however, more standard to view $L^2(X)$ to be the set of equivalence classes of functions whose domain is $X$ and are square-integrable over $X$. Two functions $f$ and ...

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As the other answers have said, your reasoning is very off: these are divergent sequences, and don't behave the way you think they do. On the other hand, a reasonable question to ask is: "Can we make sense of these divergent series?" The answer is yes, although it is not easy: there are several methods for summing (some) classically divergent series. See ...

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