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Questions tagged [measurable-functions]

For questions about measurable functions.

A measurable function is a structure preserving map between measurable spaces. This means that the inverse image of a measurable subset of the codomain is a measurable subset of the domain. It is a function between the underlying sets of two measurable spaces that preserves the structure of the spaces: the preimage of any measurable set is measurable. This is in direct analogy to the definition that a continuous function between topological spaces preserves the topological structure: the preimage of any open set is open. In real analysis, measurable functions are used in the definition of the Lebesgue integral. In probability theory, a measurable function on a probability space is known as a random variable.

1185 questions
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Prove that the set of reachable values is convex.

Let $f:[0,1] \to \mathbb R^n$ be an integrable function. Prove that the set $$\left\{\int_A f(x)\ \mathrm dx: A\subset [0,1] \text { is measurable}\right\}$$ is convex. Proving the statement ends up being equivalent to proving that $$\forall…
Kitegi
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Why is the definition of a measurable function seemingly backwards

Why not: Given a measurable space $(\Omega, \mathcal{A})$, a function $X:\Omega \to \mathbb{R}$ with Borel $\sigma$-algebra $\mathfrak{B}(\mathbb{R})$, is measurable if $\forall A \in \mathcal{A}, X(A) = B \in \mathfrak{B}(\mathbb{R})$ ? I'm…
TonyK
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How do I find an example s.t. a function is not mesurable on the completions of the simga-algebras.

I have the following problem: Let $(\Omega_1,A_1,\mu_1),(\Omega_2,A_2,\mu_2)$ be measure spaces and $f:\Omega_1\rightarrow \Omega_2$ be a $(A_1,A_2)$-mesurable map. We denote by $A_1^*,A_2^*$ the completions of the $\sigma$-algebras $A_1,A_2$.Is…
user123234
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Why does this function split up like this? (Rudin, Real and Complex Analysis)

This is regarding Theorem 1.8 on page 11 of Rudin's Real and Complex Analysis, we have a measurable space $X$ and two real valued functions on it $u$ and $v$. We then have a continuous mapping $\Phi$ from the real plane into a topological space $Y$.…
Charlie
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If $F\subset X\times Y$ is a measurable, is it true that $\chi_F$, seen as a function from $X$ to $L^\infty(Y)$ is measurable?

Let $(X,\mathcal F_{X})$ and $(X,\mathcal F_{Y})$ measurable spaces and $F\in \mathcal F_{X}\otimes\mathcal F_{Y}$. Let $L^\infty(Y)$ be the space of bounded measurable functions from $(Y,\mathcal{F}_Y)$ to $\mathbb{R}$, equipped with the sup…
Bob
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If $f:\mathbb{R}\to\mathbb{R}$ measurable with $f(x) \neq 0$, prove there exists measurable $g(x)$ with $\frac{1}{2}f(x) < g(x) < f(x)$

The question I have been assigned is this: If $f:\mathbb{R}\to\mathbb{R}$ measurable with $f(x) \neq 0, \forall x \in \mathbb{R}$, prove there exists measurable $g(x)$ such that $\frac{1}{2}f(x) < g(x) < f(x), \forall x\in \mathbb{R}$. My question…
Flose
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If $(X,d)$ is a metric space, is $X^{m+1}\to\{1,...,m\}, (x,x_1,...,x_m)\mapsto\min(\operatorname{argmin}_{k\in\{1,...,m\}}d(x,x_k))$ measurable?

Given a a metric space $(X,d)$ and $m\in\mathbb{N}$, equip $X\times X^m$ with the product Borel $\sigma$-algebra and $\{1,...,m\}$ with the $\sigma$-algebra $2^{\{1,...,m\}}$. Is the map $$X\times X^m\to\{1,...,m\},…
Bob
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Question: Let $f$ is continuous, $g$ is measurable function.Show that composition function $f\circ g$ is measurable

How to show that composition of $f$ and $g$ is measurable?
Shashi
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Measurable functions having different codomains

I just started studying Rudin's RCA,and I find that the codomains of measurable functions appearing in statements vary.Sometimes $[0,∞]$,sometimes$[-∞,∞]$. Since a measurable function is a function from a measurable space to a topology space in…
Mugenen
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Show if function is Lebesgue-measurable

I want to show if $f(x_1,x_2)=\dfrac1{1-x_1x_2}$ is Lebesgue-measurable or not on $[0,1)^2$. How do I start in this case, because the function is 2 variables? Normally, I would look if the set $\{f>a\}$ is in the $\sigma$-algebra $\forall$ $a$, as I…
user521181
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Measurability of Partial Integrals, for Conditional Expectation

I'm reading through a text that contains a more abstract definition of expectation than what I'm used to. The definition I'm used to is: $E(X|Y)(y)=\int xf_{X|Y}(x,y)dx$, where $f_{X|Y}$ is the conditional probability density function…
J.D.
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If a sequence of pointwise bounded functions satisfies this property then their domain has null measure

I'm trying to solve the following problem from the problem set of a Real Analysis course: Let $E \subseteq \mathbb R^m$ be a measurable set (w.r.t the Lebesgue measure $\mu$) and $(f_k)_{k \in \mathbb N}$ a sequence of measurable functions from $E$…
Francisco José Letterio
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Example of non-measurable function

Lets take Definition: Example: Then f(x) is not a measurable function, because ${f^{-1}}$({3,4}) $:=$ {1,2} $\not\in \Sigma $. Is this correct ?
kolo kolo
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What can be said about the measurability of $f(x,\cdot):Y\to\mathbb{R}$ if $f:X\times Y\to\mathbb{R}$ is not measurable?

Assume that $(X,\Sigma)$ and $(Y,\text{T})$ are measurable spaces and the function $f:X\times Y\to\mathbb{R}$ is not measurable. Can we say anything about the (non) measurability of the function $f(x,\cdot):Y\to\mathbb{R}$ for fixed $x\in X$?
Nicki
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How to show this function is measurable

This is a problem from the S.-T YAU COLLEGE MATH CONTEST in 2012. The original problem is Let $f(x)$ be a real measurable function defined on $[a,b]$. Let $n(y)$ be the number of solutions of the equation $f(x) = y$. Prove that $n(y)$ is a…
Apocalypse
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