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16 votes
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Why the characteristic function is measurable?

$\mathcal{X}_E^{-1}(\{0\}) = E^C$, which is perfectly measurable. Take any $A \in \mathcal{B}(\mathbb{R})$: $\mathcal{X}_E^{-1}(A) =\begin{cases} X, & 0,1 \in A \\ E, & 1 \in A, 0 \notin A \\...
dEmigOd's user avatar
  • 3,320
15 votes
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Increasing function is measurable

Let $S = f^{-1}([c, \infty))$. By definition, $S=\{x|f(x)\ge c\}$. Note that whenever $a\in S$ and $b>a$, then $f(b)\ge f(a) \ge c$, so $b \in S$. That is, for every number in $S$, all larger ...
Karl's user avatar
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11 votes
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Why is the definition of a measurable function seemingly backwards

With your definition the characteristic function of any set would be measurable, since the image $\{1\}$ is Borel. Now, working with the usual Vitali non-measurable set $E$ , you have a set $E\...
Martin Argerami's user avatar
10 votes
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For measurable $f: \mathbb{R} \rightarrow \mathbb{R}$ prove $f(x)$ and $\frac{1}{f(1/x)}$ cannot both be Lebesgue integrable.

This is a cute problem. Suppose both $$\int_{\mathbb R}|f(x)|dx<\infty\text{ and }\int_{\mathbb R}\frac{1}{|f(\frac 1 x)|}dx<\infty$$ hold. Note that this implies $f(x)\ne0$ for almost all $x$....
kimchi lover's user avatar
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9 votes
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Borel Measurable Function but not Lebesgue Measurable

The map $f$ is continuous, so the preimage of a Borel subset of $\mathbb{R}^2$ will be a Borel subset of $\mathbb{R}$. Thus $f$ is Borel to Borel measurable. Let $N\subseteq[0,1]$ be a Lebesgue ...
John Griffin's user avatar
  • 10.7k
9 votes
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Can the supremum of an uncountable family of measures be replaced by the supremum over a countable subfamily?

This is not true in general. For instance, if $X$ is uncountable and $\mathcal{A}$ contains all the singletons (say, $X=\mathbb{R}$ and $\mathcal{A}$ is the Borel sets) then the supremum of the delta-...
Eric Wofsey's user avatar
8 votes

Function that fails to be differentiable on a set of measure zero.

By outer regularity, there is a nested sequence of open sets $U_n$ so that $E \subseteq U_n$ and $m(U_n) < 2^{-n}$. Let $$f(x) = x + \sum_{n=1}^\infty m(U_n \cap (-\infty, x))$$ To see that $f$ ...
Robert Israel's user avatar
8 votes

Multivariable function having partial derivatives almost everywhere are necessarily measurable?

Even more general result is true: Theorem. A function $f:\mathbb R^n\to \mathbb R$ is Lebesgue measurable if $f$ is separately continuous almost everwhere. Proof. This theorem can be proved by ...
Taras Banakh's user avatar
7 votes
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Markov kernel in simple random work

I'm not sure what exactly is your question. Markov kernels are a way of specifying Markov processes, extending the concept of transition matrix for (finite-state) Markov chains. More generally, they ...
Blackbird's user avatar
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7 votes
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Suppose $f$ is a measurable function and $f(x) > 0$ for all $x$. Let $g(x) = \frac{1}{f(x)}$. Prove that g is a measurable function.

To show that a function is measurable, it is necessary to show that the preimage of each measurable set is measurable. To do this, it is sufficient to show that the preimage of each interval of the ...
Xander Henderson's user avatar
7 votes
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Constructing a Measure Space where $\mu(A) = 1$ if $A^C$ is Uncountable

(1) $\mu$ is a measure: Since $\varnothing$ is countable, then $\mu(\varnothing)=0$. Let's take $E_1,E_2, \cdots \in \mathcal{A}$ pairwise disjoint. If every $E_n$ is countable then $$ \mu\left( \...
Alonso Delfín's user avatar
7 votes
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Projection of a measurable set of positive measure.

Projections need not be measurable, but if they are measurable then their measures are necessarily positive: if $A$ and $B$ are the projections of $E$, then $E \subset A\times B$ and the measure of $A\...
Kavi Rama Murthy's user avatar
7 votes
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Understanding supremum / infimum of a sequence of functions in context of sequences of measurable functions

It helps to be pedantic here: $\sup_n f_n(x)$ is not a function: it's a number. Specifically, it's the supremum of the set $\{f_1(x), f_2(x), \ldots\}$ which you should be familiar with from pre-...
parsiad's user avatar
  • 25.2k
7 votes
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Measurability conditions in the definition of Brownian motion

First answer: We need to show that for any $\omega \in \Omega$, function $Y_\omega(t)= X(t,\omega)$ is $\mathcal B([0,\infty))$ measurable. Note that $Y_\omega = X \circ h_\omega, $ where $h_\omega(t)...
Dominik Kutek's user avatar
7 votes
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Why do we require a function to be measurable in order to define its Lebesgue Integral?

Without measurability, that definition could be called the lower Lebesgue integral $\underline{\int} f \,d\lambda$. Unfortunately, it lacks desirable properties. For example $$ \underline{\int}(f+g)\...
GEdgar's user avatar
  • 113k
6 votes
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Why isn't the approach of measurable functions (almost) duplicated in the approach to computable functions?

First of all, at a higher level I think it's worth mentioning the extremely well-developed analogy between the hyperarithmetic hierarchy of sets of natural numbers and the Borel hierarchy of sets of ...
Noah Schweber's user avatar
6 votes
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Why are pointwise measurable functions not measurable?

Yes, there is. Indeed consider the Borel measure on $[0,1]$ and let $A$ be your favorite non measurable subset of the unit interval. Then define a function $f$ on $[0,1]^2$ as follows: $$f(x,y) = \...
Giovanni's user avatar
  • 6,429
6 votes
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Does there exist a measurable function with a specific property?

It's not possible. Let $M$ be such that the set of $t$ with $|f(t)|<M$ has measure at least $1/2$. Then divide the interval $I_0=[-M,M]$ into halves at it midpoint, and choose the half $I_1$ with $...
C Monsour's user avatar
  • 8,256
6 votes
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The inverse image of a measurable set under a measurable function is measurable?

It depends what $\sigma$ algebra you are considering on the target space. When everyone talks about measureable functions on $\mathbb R$, they mean that $\mathcal O_Y$ is the the $\sigma$-algebra of ...
D_S's user avatar
  • 34.2k
6 votes
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Is the domain of convergence of a sequence of measurable functions measurable? (general targets)

This is a really instructive question: apparently, completeness is the key! First, let $X=Y=[0,1]$ with the usual topology. We define an example for a sequence of functions $f_n: X\rightarrow Y$, ...
A. Pongrácz's user avatar
  • 7,418
6 votes

If $\int_0^1f^k(x)\;dx\le M$, then $m(\{x\in[0,1]:f(x)>1\})=0$.

Suppose that $m(E)>0, E=(\{x\in[0,1]:f(x)>1\}$. Let $E_n=m(\{x\in[0,1]:f(x)>1+{1\over n}\})$, $E=\cup_nE_n$ implies that $m(E)=lim_nm(E_n)$ since $E_n\subset E_{n+1}$. This implies that there ...
Tsemo Aristide's user avatar
6 votes
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Dominated convergence theorem counterexample

Here, $\lim_{n \to \infty}f_n$ is the pointwise limit of $f_n$. So, fix $x$ & let $n \to \infty$. Does this $x$ work for all $n$? $\int_{(0,1)}f_n d\mu = \int_{(0,1)}n 1_{(0, 1/n)} d \mu= n\int_{(...
SL_MathGuy's user avatar
  • 1,584
6 votes
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Example of a Riemann-integrable function and non-Borel-measurable

Let $C$ be the Cantor set, and $D \subseteq C$ such that $D$ is NOT Borel measurable. (The construction of such $D$ can be easily found. For instance: Lebesgue measurable set that is not a Borel ...
Ramiro's user avatar
  • 17.9k
6 votes
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Example of Non-Measurable Bijection from $\mathbb R \to \mathbb R$

Take a none measurable subset $X\subset [0, 1]$ whose cardinal is the same as that of $[0, 1]$ - for example let $X$ be a non-measurable subset in $[0, \frac 12]$ union with $[\frac 12, 1]$. Let $f$ ...
Yuval's user avatar
  • 3,499
6 votes

Measurability conditions in the definition of Brownian motion

Kuo's definition of "stochastic process" is unconventional; most sources would use the term "measurable stochastic process". The measurability of the random variables $\omega\...
John Dawkins's user avatar
  • 26.5k
6 votes
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Is $f:[a,b] \times \Omega \to E$ measurable?

The the argument of the OP works fine if $E$ is assumed to be a metric space. The remaining of this posting is to show that the Lemma in the OP does not hold in general topological spaces (other than ...
Mittens's user avatar
  • 39.8k
5 votes
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How can I solve thins problem by applying the Radon-Nikodym Theorem?

Hint: Suppose that $f\geq 0$. Let $\nu$ be the measure defined on $(X,M)$ by $$ \nu(E):=\int_E f\,d\mu. $$ Now, let $\mu_0=\mu\vert_{M_0}$ and $\nu_0=\nu\vert_{M_0}$ be the restriction of these $...
Nick Peterson's user avatar
5 votes

Real analysis: measurable function $f$ and non measurable function $g$ such that their composition is measurable

An uninteresting example: $f=0$ and $g=\chi_{E}$, where $E$ is any non-measurable set, then $f\circ g=0$.
user284331's user avatar
  • 55.7k
5 votes
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A Non-Counterexample to the Fubini Theorem with Counting and Lebesgue Measures

@Diesirae92 explains the measurability and lack of contradiction; we confine ourselves to calculations. In the first inner integral, $$ \int_Y \chi_D(x,y) \, \mu(dy) $$ $\chi_D(x,y) \, \mu(dy)$ ...
Integrand's user avatar
  • 8,387

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