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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.

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Are Monotone functions Borel Measurable?

Could you guide me how to prove that any monotone function from $R\rightarrow R$ is Borel measurable? Since monotone functions are continuous away from countably many points, would that be helpful in proving the measurability?
user48405
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Are continuous functions strongly measurable?

Measure theory is still quite new to me, and I'm a bit confused about the following. Suppose we have a continuous function $f: I \rightarrow X$, where $I \subset \mathbb{R}$ is a closed interval and $X$ is a Banach space. I can show that $f$ is…
ScroogeMcDuck
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liminf and limsup with characteristic (indicator) function

So first let me state my homework problem: Let $X$ be a set, let $\{A_k\}$ be a sequence of subsets of $X$, let $B = \bigcup_{n=1}^{+\infty} \bigcap_{k=n}^{+\infty} A_k$, and let $C = \bigcap_{n=1}^{+\infty} \bigcup_{k=n}^{+\infty} A_k$. Show that…
nate
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Forward characterization of measurable functions?

In topology, the standard definition of continuity works in the "backward" direction, since it puts a condition on the pre-images of a function rather than images: $f$ is continuous if the pre-image under $f$ of every open set is open. However,…
WillG
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$f$ a real, continuous function, is it measurable?

Let $f: \mathbb{R} \to \mathbb{R} $ be a continuous function. I need to show that is a measurable function. I tried working with the definition: Let $f: X \to \mathbb{R}$ be a function. If $f^{-1}(O)$ is a measurable set for every open subset $O$…
user54297
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$f :\mathbb R \to \mathbb R$ be a bijective Lebesgue measurable function , then is $f^{-1}:\mathbb R \to \mathbb R$ Lebesgue measurable?

Let $f :\mathbb R \to \mathbb R$ be a bijective Lebesgue measurable function , then is $f^{-1}:\mathbb R \to \mathbb R$ Lebesgue measurable ? I don't think this is true but I can't find any counterexample , or any way through . Please help . Thanks…
user228168
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Definition of a measurable function?

Are these two definitions of a real-valued, measurable function equivalent? ($(X, \Sigma, \mu)$ is the measure space.) Definition 1: $f: X \to \mathbb{R}$ is said to be measurable if for all $\alpha \in \mathbb{R}$, $\{ x \mid f(x) > \alpha \} \in…
layman
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Can the supremum of an uncountable family of measures be replaced by the supremum over a countable subfamily?

Consider a measurable space $(X,\mathcal{A})$. Let $\mathcal{M}$ denote the family of all countably additive measures $\mu\colon \mathcal{A}\to [0,+\infty]$. This family can be made into a partially ordered set by setting $\mu\leq \nu$ iff…
Rafael
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Let $f : \mathbb{R} \to \mathbb{R}$ be measurable and let $Z = {\{x : f'(x)=0}\}$. Prove that $λ(f(Z)) = 0$.

The following is an exercise from Bruckner's Real Analysis: Let $f : \mathbb{R} \to \mathbb{R}$ be measurable and let $Z = {\{x : f'(x)=0}\}$. Prove that $λ(f(Z)) = 0$. For the case $f$ being nondecreasing / nonincreasing, we can defined…
user200918
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Is it possible to redefine measurable spaces such that measurable functions are not defined in terms of pre-image?

We know that a measurable space is defined as a tuple $(X,\mathcal A)$, where $X$ is a set and $\mathcal A$ is a $\sigma$-algebra defined on $X$. A $\sigma$-algebra is traditionally defined as a collection of subsets such that the whole set is in…
HackR
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$\exists f:\mathbb{R}\to\mathbb{R}$ such that $\frac{f(X_1)}{X_1+X_2}$ is independent of $X_1$?

Let $X_1, X_2$ be independent real random variables. Does there exist a non-zero measurable function $f:\mathbb{R}\to\mathbb{R}$ such that $$\frac{f(X_1)}{X_1+X_2}\text{ is independent of }X_1?$$ I am especially interested in the case when $X_1,…
Albert Paradek
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Is $f:[a,b] \times \Omega \to E$ measurable?

Problem: Suppose that $[a,b] \subset \mathbb{R}$, $(\Omega, \mathcal{F})$ is a measure space and $E$ is a topological space. Suppose $f : [a,b] \times \Omega \to E$ is such that: $\forall t$ $f(t, \cdot): \Omega \to E$ is measurable. (Here the…
Filippo Giovagnini
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Conceptual Issues in the Measure Theoretic Proof of Conditional Expectations (via Radon-Nikodym)

I have been looking into measure theory (from a probabilist's perspective), and I have found the proof of the existence of the conditional expectation to feel a little "glossed over" in literature. As such I've tried to very, very slowly break down…
tisPrimeTime
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Is the set of measurable functions measurable?

Let $M$ be the set of measurable functions in $\mathbb{R^R}$. Now, $\mathbb{R^R}$ has the borelian sigma-algebra associated with its product topology, which allows us to ask the following question : Is $M$ measurable? On the positive side, I have…
Moinsdeuxcat
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Is the domain of convergence of a sequence of measurable functions measurable? (general targets)

Let $(X,\Sigma)$ be a measurable space, and let $Y$ be a topological space. Let $f_n:X \to Y$ be a measurable sequence, when we take on $Y$ the Borel sigma algebra. In other words the $f_n$ are measurable as mappings between the measurable…
Asaf Shachar
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