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Questions about abstract measure and Lebesgue integral theory. Also concerns such properties as measurability of maps and sets.
6
votes
Accepted
Relation between measure of sets
The answer is no. Consider $B\subset A$ with $\mu(B)$ very small, $C_1=A\setminus B$ and $C_2=X\setminus C_1$. Then the r.h.s. equals $\mu(B)^2/\mu(A)\mu(C_2)$ which can be less than $\mu(B)$ since $ …
20
votes
1
answer
907
views
A collection of intervals that can cover any measure zero set
This is a follow-up to this question (in fact, this is what originally motivated me to ask that one.)
Let's say that a sequence $\{s_i\}$ of positive reals covers a set $X\subset\mathbb R$ if there i …
27
votes
2
answers
2k
views
A set that can be covered by arbitrarily small intervals
Let $X$ be a subset of the real line and $S=\{s_i\}$ an infinite sequence of positive numbers. Let me say that $X$ is $S$-small if there is a collection $\{I_i\}$ of intervals such that the length of …
30
votes
Accepted
Zariski closed sets in C^n are of measure 0
If a real analytic function $f:U\subset\mathbb R^n\to\mathbb R^m$ is zero on a set $Z$ of positive measure (and $U$ is connected), then $f\equiv 0$.
Indeed, almost every point of $Z$ is a density poin …
29
votes
Why is Lebesgue integration taught using positive and negative parts of functions?
The integration of nonnegative functions deserves its own chapter, just like nonnegative measures. It has more features than the general case and there are cases when you need exactly these features a …
33
votes
1
answer
4k
views
Is every smooth function Lebesgue-Lebesgue measurable?
This is motivated by pure curiosity (triggered by this question). A map $f:\mathbb R^n\to\mathbb R^m$ is said to be Lebesgue-Lebesgue measurable if the pre-image of any Lebesgue-measurable subset of $ …
8
votes
1
answer
380
views
Estimating flat norm distance from a planar disc
Let $D\subset\mathbb R^2\subset\mathbb R^n$ be a unit planar disc in $\mathbb R^n$. Let $S$ be an orientable two-dimensional surface in $\mathbb R^n$ such that $\partial S=\partial D$. Of course, we h …