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1

Your objection is indeed correct Here is another demonstration: $$\int_{\Bbb R} e^{-ixy}f'(x) dx = \int_{\Bbb R} \lim_{h\to 0} e^{-ixy}\frac{f(x+h)-f(x)}{h} dx$$ $$= \lim_{h\to 0} \frac{1}{h}\int_{\Bbb R} e^{-ixy} (f(x+h)-f(x))dx$$ $$= \lim_{h\to 0} \frac{1}{h}\int_{\Bbb R} e^{-i(u-h)y}f(u)du - \hat{f}(y)$$ $$= \lim_{h\to 0} \frac{e^{ihy}-1}{h} ... 1 It's not necessary to prove that \lim_{n\to\infty}\mathbb{P}(X_n\leq c)=0. Instead, let$$A=\{\alpha\in\mathbb{R}:\lim_{n\to\infty}\mathbb{P}(X_n\leq \alpha)=0\} $$and observe that if \alpha<c, then there exists \gamma>0 such that \alpha<c-\gamma<c, hence$$\mathbb{P}(X_n\leq \alpha)\leq \mathbb{P}(X_n<c-\gamma)$$for all n, so$$ ...

0

If $\mu$ is a Radon measure (which is not an inappropriate hypothesis for a measure on a locally compact Hausdorff space) then the assertion is true. To see this fix $\epsilon>0$ and define $B:=\{x\in X: g(x)>\epsilon\}$. Because $\mu$ is Radon there is a sequence $\{K_n\}$ of compact subsets of $B$ with $\sup_n\mu(K_n)=\mu(B)$. Because $K_n$ is ...

3

If $X$ is not $\sigma$-finite, the answer is NO. Here is a simple example. Let $X=[0,1]$ with the usual topology. Let $\mu$ be defined on the Borel $\sigma$-algebra, by $\mu(\emptyset)=0$ and $\mu(E)=+\infty$ if $E\neq \emptyset$. It is easy to see that$\mu$ is a regular measure. Let $g$ be the constant function, $g=1$. We have $\int_E g = 0$ for all ...

0

The proof depends on the form of the minimizer $g$: $$g(x) := \chi_{(f<s)}(x) + c\chi_{(f=s)}(x).$$ For any $h$ satisfying $0\le h\le1$ and $\int_\Omega h\,d\mu=G$, we compute \begin{equation*} \begin{split} \int_\Omega fg\,d\mu &= \int_{\{f<s\}}f\,d\mu + c\int_{\{f=s\}} f\,d\mu \\ &= \int_{\{f<s\}}f\,d\mu + cs\mu(\{f=s\}) \\ &= ...

1

$\text{“}E_i$ and $F_i$ are disjoint$\text{''}$ could be construed to mean $E_i\cap F_i=\varnothing$, and that is not true. It is true that $E_1,E_2,E_3,\ldots$ are pairwise disjoint. Suppose $x\in\bigcup_i F_i$. Then there is some smallest index $i_0$ such that $x\in F_i$. For that smallest index $i_0$ we have $x\in E_{i_0}$; therefore $x\in\bigcup_i ... 0 Continuous functions are not always measurable. It depends on the$\sigma$-algebra! Actually, continuous functions are measurable for an algebra$\mathcal{F}$if and only if$\mathcal{F}$contains the Borel algebra. I think you can see why it is sufficient. To see that it is necessary, suppose that all continuous functions are$\mathcal{F}$-measurable, then ... 0 As @Did pointed out, there was an error in my answer in the first part. The following is my new attempt (though not complete). We have$\sum\limits_{n \leq N} 1_{E_n} \geq \frac{\epsilon N}{2} $iff there are at least$K$(and at most$N$) numbers$1$, where$K = 1+$integer part of$\frac{\epsilon N}{2}$(if$\frac{\epsilon N}{2}$is integer then there ... 0 Of course, as noted in the other answers,$\Omega\in\mathcal{F}$for any$\sigma$-algebra. However, I will try to spell out the calculation with more detail. If$X$is$\mathcal{G}$-measurable, and$\mathcal{F}\subset\mathcal{G}$, then for any$A\in\mathcal{F}$,$E(\mathbf{1}_AE(X\mid \mathcal{F})=E(\mathbf{1}_AX)$. That's the definition of conditional ... 0 What I have read (I believe in "A Modern Approach to Probability Theory" by Fristedt and Gray) is that one of the primary reasons for using Riemann-Stieltjes integration is the convenient integration by parts formula that comes with it. However, the Riemann-Stieltjes integral suffers several of the same problems as the Riemann integral which the Lebesgue ... 1 It's false. (I was taking a shower when Michael posted his comment; what's below is a detailed exposition of a simpler version of what he said.) Say$G$is the group$[0,1)$, with addition modulo$1$. Note that Lebesgue outer measure is$G$-invariant. I'll be writing$a+b$for the addition in$G$. Let$H=[0,1)\cap\Bbb Q$, and let$C$be a complete set of ... 3 If$[h \sim c]$denotes the set$\{x\mid h(x) \sim c\}$(where$\sim$is any relation), notice that your set is: $$[f - g = 2] = [f - g \le 2] \cap [f - g \ge 2]$$ both of which are measurable since$f - g$is measurable. 4 For 1, just note that it's a telescoping sum. That is: $$\sum_{n=1}^k (nx^{n-1}-(n+1)x^n) =(1x^{1-1}-(1+1)x^1)+\cdots+(kx^{k-1}-(k+1)x^k)=1-(k+1)x^k$$ Just take the limit as$k \to \infty$and you get$1$For 2, just do the integral, The antiderivative is just the power rule. 1 You are correct. You could also say that$\emptyset \subseteq \Bbb{R}$and every$\sigma$algebra contains$\emptyset$. However, if$A\neq \Bbb{R}$and$A\neq \emptyset$then just consider the trivial$\sigma$algebra$\{\Bbb{R},\emptyset\}$. 0 although not stated this way in books ,I believe it is true after reading page 98 of 'The Integrals of Lebesgue, Denjoy , Perron , and Henstock (Graduate Studies in Mathematics Volume 4 ) by Russell A. Gordon ' LEMMA 6.14. Let F : [a, b] --> R be measurable and let E subset of [a, b] be measurable. If F is differentiable at each point of E, then µ* (F (E)) ... 1 You acknowledge that there is no mystery in the definition of p-adic measures attached to a profinite abelian group$G$, which are simply elements of the complete group algebra$\Lambda(G)$. As for p-adic pseudo-measures,they are elements of the total ring of fractions$Q(G)$satisfying a certain technical condition described at the bottom of p.35 of the ... 1 First let's show that$A$is Borel. Given$\epsilon>0$, let$A_\epsilon=\left\{h\in[0,1]:\exists x\in[0,1-h]\text{ such that }|f(x)-f(x+h)|<\epsilon\right\}$. Since$f$is continuous, we can consider only rational numbers in the definition of$A_\epsilon$, that is, if$\left\{q_n:n=1,2,\ldots\right\}$denote the rationals in$[0,1]$, we have ... 1 In general no: simple functions are dense but you cannot "hold fixed" the value on$A_1$as you increase the number of sets. This is much like how polynomials are dense in the continuous functions on a compact interval, and yet not every continuous function has a power series. 4 Yes the reverse holds as well. This is really a definitional convention. We typically define "$f$is integrable" to mean that both$\int_\Omega f^+ d\mu$and$\int_\Omega f^- d\mu$are finite (since, in the beginning, we define integration only for non-negative functions) then define $$\int_\Omega f d\mu = \int_\Omega f^+ d\mu - \int_\Omega f^- d\mu.$$ ... 0 Let$\lambda$be the Lebesgue measure on$\mathbb{R}^n$, and$\mathcal{L}$the Lebesgue$\sigma$-algebra. Take a Vitali set$V\subseteq[0,1]^n$(i.e., a set of representatives of$\mathbb{R}^n/\mathbb{Q}^n$), which is non-Lebesgue-measurable for$\lambda$and has cardinality$|V|=|\mathbb{R}^n|=\mathfrak{c}$. Take a nonempty open set$B\subseteq(2,3)^n$, ... 0 This property is clearly true for non-negative simple functions (since simple functions are bounded). Fix$\epsilon>0$. By the construction of the integral, there exists a bounded function$g$such that$g\leq |f|$and$\int |f| d\mu< \int g d\mu +\epsilon/2$. This means that for any subset$E$, $$\int_{E} |f| d \mu - \int_{E} g d \mu \leq \int |f| d ... 3 Note that X is a finite set. So given x\in X, the set B_x= \{ S \in \mathcal{F} : x\in S\} is finite. Since \mathcal{F} is a \sigma-algebra, we have X\in B_x (B_x is not empty) and$$A_x=\bigcap_{S\in B_x} S \in \mathcal{F}$$Clearly x\in A_x, and A_x is the smallest element in \mathcal{F} containing \{x\}. Note that if y\in X and ... 2 Take$$S_n=\sum_{k=1}^n\left|f_k\right|$$and apply the monotone convergence theorem for \left\{S_n\right\}_{n=1}^\infty. 2 To be true that A_i \downarrow A \implies \mu(A_i) \downarrow \mu(A), you should have that \mu(A_1) < \infty. Hint: construct an example where \mu(A_1) = \infty. 2 Try this, which I think is similar to your strategy. Let C=\{x_1, x_2, \cdots \} be a countably infinite set of distinct points. Let A_n:= \{x_i \in C : i\geq n\}. So A_1=C, A_2=\{x_2, x_3,\cdots \}, etc. Then Clearly$$\bigcap_n A_n =\emptyset$$and A_n \downarrow. However, for each n, \mu(A_n)=\infty, so \mu(A_n) cannot tend to 0. 2 There is a set A \subset \Omega that can be identified as \mathbb{Z} (in the sense that there is a bijection f: \mathbb{Z} \to A). Let A_n = \mathbb{Z} \setminus \left\{ 1 ,\dots, n \right\}. This sequence satisfies the claim. 2 Choose x_k \in \Omega such that each x_k is distinct. Let A_n = \{x_k\}_{ k \ge n}. 2 Let A_n be the natural numbers with the first n natural numbers thrown out. I.e. A_n=\mathbb N\setminus\{1,2\cdots n \}. Then each \mu(A_n)=\infty and \cap A_n=\phi so that \mu(\cap A_n)=0. Can you adapt this argument to your example? 0 Hint: Note that [a,\infty)\cap (\Bbb{R}\setminus [b,\infty))= [a,\infty)\cap(-\infty,b) = [a,b) if a< b in \Bbb{R}. Then f^{-1}([a,b))=? 0 You don't need to prove \lim_{n \to \infty}f_n(x) = f(x) . We can prove directly that \liminf_{n \to \infty} \int_{U}f_n(x)dx \geq \int_{U}f(x)dx\: for any open set U of \mathbb{R}. Here is, in detail, a simple way to prove it. Since f_1, f_2, \cdots and f are integrable functions on \mathbb{R}, we have for all y \in ... 2 In the paragraph above Lemma 2.6, the authors explicitly state "[we write] \mathcal L^\infty for the space of measurable bounded functions" 3 Let me mention some related results. Given a compact metric space X, the set Y of all Borel probability measures on X is metrizable and in fact the induced topology on Y makes it compact. The first property is a more or less easy consequence of the separability of the space C(X) of continuous functions on X with the supremum norm, which allows ... -2 Let f: [a,b]\rightarrow \mathbb{R} is measurable. Suppose A=\{x: f(x)\neq 0\} is of positive measure.. With out loss of generality and by abuse of notation, assume that A=\{x:f(x)>0\} is of positive measure Again by With out loss of generality and by abuse of notation, assume that A=\{x:f(x)>\epsilon\} is of positive measure. So, there ... 0 I haven't worked out the details - also I suspect we still haven't been given all the relevant definitions. But in case it helps, here's how the argument "must" go in outline, if it's by R-N: Somehow we reduce to the case \mu(\Omega)<\infty. Define a complex measure \nu by$$\nu(E)=x^*(\chi_E).$$Detail: Something shows somehow that \nu is in fact ... 1 Let c \in \Bbb R and let [h<c] denote the set \{x \mid h(x) < c\}. x \in [g_a<c] \iff g_a(x) < c \iff f(x+a) < c \iff x+a \in [f < c] \iff x \in [f<c] - a So, [g_a<c] = [f<c] - a. But since the Lebesgue \sigma-algebra \Bbb L is closed under translations and [f<c] \in \Bbb L, we get [f<c] - a\in \Bbb L, so ... 0 On B_n, estimate |f|^2 by n|f|, the n is cancelled out by the inner sum, you remain with \sum_n\int_{B_n}|f|=\int_{E}|f|<\infty. 0 In every metric space (X,d) a sequence x_n converges to x if and only if every subsequence x_{n_k} has a further subsequence converging to x (easy proof by contradiction). All you need to now is thus that convergence in measure is convergence in a metric space (e.g. d(f,g)=\int \min\lbrace 1,|f(x)-g(x)|\rbrace \, d\mu(x) is a suitable metric on ... 0 If \mu(X)<\infty, then this follows from Jensen's inequality (\because x\mapsto\lVert x\rVert is convex), i.e.$$ [\mu(X)]^{-1}\left\lVert\int f d\mu\right\rVert=\left\lVert [\mu(X)]^{-1}\int f d\mu\right\rVert \le [\mu(X)]^{-1}\int \lVert f\rVert d\mu. $$1 Think of |f| as a division of f into two functions: f_+ and f_-. f_+ we define as equal to f on the domain {x: f(x) is non-negative}, and 0 on all other x. f_- we define as equal to -f on the domain {x: f(x) is negative} and 0 elsewhere. |f|=f_+ + f_-. If |f| is finite, then necessarily both f_+ and f_- are finite. 3 No, it does not: If E is (bounded and) non-measurable, then f = 2\chi_{E} - 1 is everywhere equal to \pm1, so |f| \equiv 1, but f is non-measurable, hence not integrable. 1 Assume f to be measurable. f is bounded by |f| and so it is L^1, and hence Lebesgue integrable. On the other hand, f is lebesgue iff \int_{X}f^+ \, d\mu<\infty and \int_{X}f^- \, d\mu<\infty, but we know that |f|=|f^+ + f^-| \leq |f^+|+|f^-| and so |f| is integrable if |f| is. 2 By the monotone convergence theorem, you always will have$$\int f^p = \lim_{k \rightarrow \infty} \int f_k^p$$Taking pth roots (and using continuity of the function x \rightarrow x^{1 \over p}), one therefore has$$||f||_p = \lim_{k \rightarrow \infty} ||f_k||_p \tag 1$$Since the f_k increase to f, the limit in (1) is an increasing limit. So one ... -1 Oh, I just read your updated version of the question. If fk->f in Lp, we know that f is in Lp. Considering fk is a non-negative increasing sequence of functions, to converge at f it must be that sup(k) ||fk||p=||f||p. f is in Lp, so ||f||p is finite. 0 If f_k\to f in L^{\infty}, then for each n\in\mathbb{N} there is an index k_n such that ||f-f_k||_{\infty}<\frac{1}{n} for all k\geq k_n, hence for each k\geq k_n there is a nullset N_{k,n} such that$$ \sup_{X\setminus N_{k,n}}|f_k(x)-f(x)|<\frac{1}{n} $$Let$$N=\bigcup_{n=1}^{\infty}\bigcup_{k=k_n}^{\infty}N_{k,n}$$then$N$is a ... -1 Well, fk is a sequence of increasing functions to f, so with sufficient work you should be able to show that sup(k) ||fk||p = ||f||p. 0 Take$g=1_{\{f\geq0\}}-1_{\{f<0\}}$to get$\int fgd\mu=\int|f|d\mu=\Vert f\Vert_1$. Note that$g$is measurable if$f$is. (Also you might want to change$\sup$to$\operatorname{esssup}$in your use of$\Vert\cdot\Vert_\infty$) 0 You already have the right answer. The case of$\alpha=0$is trivial. Consider$\alpha\neq0. Then (being a bit loose; you can fill in the details), \begin{align*} \left\Vert \alpha f\right\Vert & =\inf\left\{ M\geq0\colon\left|\alpha f(x)\right|\leq M\text{ for almost all }x\in E\right\} \\ & =\inf\left\{ ... 2 Your inequality\displaystyle \left| \int f \, d\mu \right| \le \int |f| d\mu$on the last line isn't correct. What if$\mu$is a negative measure? The Hahn-Jordan decomposition is (essentially) unique, so your equation 1. may as well be the definition of the integral with respect to the signed measure$\mu$. (If it isn't what is the definition you are ... 1 The version on wikipedia is wrong (if it's exactly as you say; a link might have been appropriate). This is your chance to do a Good Thing by finding the Edit button and fixing it. Counterexample with$\mu$finite but$\nu$not$\sigma$-finite: Let$X=\{0\}$. Define$\mu(X)=1$,$\nu(X)=\infty$. That was easy. May as well mention that it's just as easy to ... 2 Define$g_M := \max(\min(g,M),-M) ∈ L^1 ∩ L^\infty$and apply your argument modified so that$s_k → \operatorname{sgn} g_M$in$L^1$to find that$‖g_M‖_1 = 0$. On the other hand,$|g_M| \uparrow |g| ∈ L^1$, so MCT says$∫|g_M| → ∫ |g|$, which means$∫|g| = 0$. (Actually since$g_M^2 \uparrow g^2$as well you don't need to modify your$s_k\$ but I'll just ...

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