New answers tagged

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 $p$th 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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