# Tagged Questions

Questions related to measures, sigma-algebras, measure spaces, Lebesgue integration and the like.

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### Continuity and convergence everywhere.

Suppose a sequence of continuous functions $(h_n)$ converges almost everywhere to another continuous function $h$ . Is it possible to infer that $h_n$ infact converges everywhere? If not, under what ...
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### Question with Lebesgue-Stieltges outer measure

Let $g$ and $h$ be two increasing functions and $\theta_g$, $\theta_h$ be the associated Lebesgue-Stieltges outer measures on $R$ (the set of real numbers). We can also associate to $g+h$ the L-S ...
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### Real Analysis, Folland Proposition 3.11 The Radon Nikodym Theorem

Proposition 3.11: If $\mu_1,\ldots,\mu_n$ are measures on $(X,M)$, there is a measure $\mu$ such that $\mu_j\ll\mu$ for all $j$ -- namely $\mu = \sum_1^n\mu_j$. Attempted proof: Suppose we have a ...
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### Real Analysis, Folland Corollary 3.10 The Lebesgue Radon Nikodym Theorem

Background Information: Proposition 3.9 - Suppose that $\nu$ is a $\sigma$-finite measure and $\lambda$ are $\sigma$-finite measures on $(X,M)$ such that $\nu\ll \mu$ and $\mu\ll \lambda$. a.)...
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### Dealing with a Sequence of Sets with Two Indices and Simple Function based on that Sequence of Sets

Okay so I have a measurable function $f$ and a set $E_{n,i}$ $$E_{n,i}=\left\{ x:\frac{i-1}{2^{n}} \leq f(x)<\frac{i}{2^{n}}\right\}$$ where $i=1,...,n2^{n}$ and $n=1,2,...$ Then I have another ...
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### $p$-power integral and $p$-series in higher dimensions

This seems like a basic question that should be addressed in a multivariable calculus course however I don't think I've ever confronted the issue until I became confused about a recent question here ...
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### Is diameter of a set a measure?

Suppose the diameter of a nonempty set $A$ is defined as $$\sigma(A) := \sup_{x,y \in A} d(x,y)$$ where $d(x,y)$ is a metric. Is $\sigma(.)$ a 'measurement'? I.e., how do I prove the countable ...
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### A function is Borel iff $f^{-1}(a,\infty) \in \mathcal{F}$

I'm reading Shao's Mathematical Statistics and part of a proposition is that, if $(\Omega, \mathcal{F})$ is a measurable space then: A function $f$ is Borel iff $f^{-1}(a,\infty)\in \mathcal{F}$ for ...
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### How do you rigorously explain the fact that $u \in L^p$ can be non defined over sets of measure 0?

In all the definitions of $L^p(\Omega)$ spaces I have been given these are defined to be the set of functions $f: \Omega \to \mathbb{R}$ whose norm $||\cdot||_{L^p}$ is finite. We define is as the ...
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### Cardinality of a borelian

My advisor told me that a Borel set can only be finite, countably infinite or having the cardinality of the continuum (obviously we are not assuming Continuum Hypothesis). I think he mentioned "...
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### Showing region between $x$ axis and graph is in ${\cal{B}}(\mathbb{R^2})$

Let $(\mathbb{R^2},{\cal{B}}(\mathbb{R^2}),\lambda^2)$ be a measure space where $\lambda^2$ is the two dimensional Lebesgue measure. Let $u:\mathbb{R}\to [0,\infty]$ be a Borel measurable function. ...
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### Inner lebesgue measure definitions

I´m currently studying lebesgue measure theory. I'm using Lebesgue integration on euclidean space by Frank Jones as one of my reference books. He defines the outer and inner lebesgue measures for an ...
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### Example of set, finite outer measure, subsets, where outer measure does not converge

What is an example of a set $X$ and a finite outer measure $\mu^*$ on $X$, subsets $A_n \uparrow A$ of $X$, and subsets $B_n \downarrow B$ of $X$ such that $\mu^*(A_n)$ does not converge to $\mu^*(A)$ ...
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### Does $\mu^*$ agree with $\mu$, measure space? [closed]

If $(X, \mathcal{A}, \mu)$ is a measure space, define$$\mu^*(A) = \inf\{\mu(B) : A \subset B,\,B \in \mathcal{A}\}$$for all subsets $A$ of $X$. I have a few questions? Is $\mu^*$ an outer measure? ...
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### JL Doob / KL Chung paper: Fields Optionality and Measurability

I was reading the following paper: https://www.jstor.org/stable/2373011?seq=1#page_scan_tab_contents I had a question about a proof that is part of the paper. Here are some of the definitions and ...
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### Prove that if $X$ and $Y$ are independent, then $h(X)$ and $g(Y)$ are independent in BASIC probability — can we use double integration?

In advanced probability we can just say: \begin{align} & P(h(X) \in A, g(Y) \in B) \\[6pt] = {} & P(X \in h^{-1}(A), Y \in g^{-1}(B)) \\[6pt] = {} & P(X \in h^{-1}(A)) P(Y \in g^{-1}(B)) \...
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### Showing that $\{(x,x):x\in [0,1]\}$ is in $\cal{B}([0,1])\otimes\cal{B}([0,1])$

I am having a hard time showing that $\{(x,x):x\in [0,1]\}$ is in $\cal{B}([0,1])\otimes\cal{B}([0,1])$. ${\cal{B}}([0,1]) = \{[0,1]\cap B: B \in \cal{B}(\mathbb{R})\}$. Any hints would be ...
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### Non-invertible measure preserving transformations of $\mathbb{R}^n$

I am looking for particular examples of measure-preserving transformations of $\mathbb{R}^n$ (with Lebesgue measure) to get a better idea of how they behave. A large family of such transformations ...
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### Showing the set of real values for which the pre-image has measure greater than zero is measure zero

The question is stated as follows: Show that if $f: \mathbb{R} \rightarrow \mathbb{R}$ is measurable, then the set $E = \{x \in \mathbb{R} \ | \ m(f^{-1}(x)) > 0 \}$ has measure zero. This ...
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### $\int_X f^p d\mu = p\int_{[0,+\infty)} t^{p-1}\mu(\{x\in X: f(x)>t\}) d\mu_t$ for any natural $p\ge 1$

Let $f:X\mapsto[0,+\infty)$ be a non-negative measurable function defined on the space $X$, endowed with the complete $\sigma$-additive, $\sigma$-finite, measure $\mu$ defined on the $\sigma$-algebra ...
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### Is there a way to write conditional expectation as an integral?

Let $E[X|G]$ be a random variable that is G-measurable and satisfies the partial averaging property, then we know that $E[X|G]$ is a conditional expectation. This is the definition I saw from Shreve's ...
### Showing that $\{x:u(x)\le v(x)\}$ is measurable where $u,v$ are measurable
Let $u,v$ be measurable functions from a measurable space $(X,\cal{A})$ to $(\mathbb{\bar{R}},\cal{B}(\mathbb{\bar{R}}))$. How can I prove that $\{x:u(x)\le v(x)\}$ is in $\cal{A}$? It is of course ...