New answers tagged lebesgue-measure
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Open and dense is full Lebesgue measure?
No. Here's an example much simpler than a fat Cantor set. Let $\{a_1, a_2, \cdots, a_n, \cdots\}$ be the set of all rational numbers in $(0,1)$. For each $i$, take a neighborhood of length $\frac{1}{2^...
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Why regularity is important for a Borel measure?
The main theorem about Borel measures is the Riesz--Markov--Kakutani representation theorem. There are several versions of it, so I am not going to be too precise here. Actually all this answer is ...
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Show that Lebesgue measure is translation-invariant.
Here is a brief proof of the one dimensional case.
Theorem: If $A \subset \Bbb{R}$ is Lebesgue measurable, then for every $x \in \Bbb{R}$, $m(A)=m(A+x)$.
Proof:
If $A \subset \Bbb{R}$ is Lebesgue ...
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Comparison between the Lebesgue integral of a given function over two different sets.
No, it is not true. You would additionally need that $m(A) \leq m(B)$. For a counterexample, consider $f \equiv 1$ and $A=[0,1]$, $B=\{2\}$.
Edit: with the additional assumption, we can prove the ...
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The upper limit of a continuous function on two variables is measurable
Verify that $\{x: F(x)<a\}=\bigcup_k \bigcup_n \bigcap_{y\in \mathbb Q, y>n} \{(x: f(x,y) \leq a-\frac 1 k\}$/
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Bochner integrability of mappings of Bochner integrable functions
If $X$ is separable, then joint measurability (inverse images of all balls are measurable) is sufficient. This is a property of Bochner measurability
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Are all measures on $\mathbb{R}$ Lebesgue-Stieltjes measures on all measureable sets?
Yes, it is true. The main idea is to use Caratheodory extension theorem. But we need to check $\ell$-finiteness. (We can check this easily because the union of $K_i$ is finite since $\mu(K)$ is finite ...
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Simple question that relates Lebesgue measure and Lebesgue integral.
It depends on how you defined the Lebesgue integral, but if you started with integrals of simple functions (as in linear combinations of indicator functions of measurable sets), then your equality is ...
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A key lemma in the proof Besicovitch Covering lemma.
As mentioned in the comments, the constant $K(n)$ needs to be independent of any bound on the radii for the family of balls, in order for the claim to be of any use in the Besicovitch covering lemma.
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Different definitions of the canonical form of simple functions (issue with constant reals different from $0$)
It seems to me that the first definition adds the restriction that $r_k$ are non-zero to make the decomposition unique. If, say, $r_{N+1}$ is allowed to be $0$, one could add it to the sum or discard ...
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Riemann integrability of a charactersitic function
Your observation regarding points of dis-continuity is not correct. For a simple example, consider the characteristic function of $\bigcup_n (\frac 1 {2n}, \frac 1 {2n-1})$. This function is not ...
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Different definitions of the canonical form of simple functions (issue with constant reals different from $0$)
I think that this restriction it for convenience: if you endow the ambient space with some infinite measure, you just have to say that all the $E_k$ must have finite measure for $\varphi$ to be ...
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A problem on the convergence of set Lebesgue measures
No. In any dimension, let $f$ be any everywhere-negative $L^1$ function, and for each $k$ define
$$
f_k(x)=
\begin{cases}
f(x) & \|x\|\leq k\\
-f(x) & \|x\|> k
\end{cases}
$$
Then $\|f_k-f\...
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$\| f \|_{L^1} > 0 $ is stronger than simply $f \neq 0$. But how much?
Yes; if $\|f\|_p>0$ then $f>0$ on some positive measure set, since otherwise $f=0$ a.e. and hence $\|f\|_p = 0$. Conversely, if $\|f\|_p = 0$ then $f=0$ almost everywhere.
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Product of a $L^{p}$ function and measure
Your intuition is correct. You cannot multiply a 'function' in $L^p(\nu)$ with a measure $\mu$ unless $\nu$ is absolutely continuous w.r.t. $\mu$.
This is independent of summability issues and just a ...
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Retrieving properties of a function $p(\cdot)$ given that $\text{ess} \sup p(x) < \infty$ or $\text{ess} \inf p(x) = \infty.$
By definition $$p_{-}=\sup\{a\,:\,\mu\{x\,:\, p(x)<a\}=0$$ If $p_{-}=\infty$ then for every $a$ there holds $\mu\{x\,:\, p(x)<a\}=0.$ We have $$\{x\,:\, p(x)<\infty\}=\bigcup_{n=1}^\infty \{x\...
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Is a set of representatives of $\Bbb R/\Bbb Z[\frac12]$ a Vitali set?
Every set of representatives of $\Bbb R/\Bbb Z[\frac12]$ will contain a pair of elements differing by a rational not in $\Bbb Z[\frac12]$ e.g. $\frac13$ and therefore it cannot be a Vitali set.
...
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Find Lebesgue measurable sets such that the measure of intersection is product of measures.
Denote the partition of $(0, 1)$ into $2^n$ equal intervals by $P_n$. Let $A_n$ be the union of the odd numbered intervals in $P_n$. The Lebesgue measure of $A_n$ is $1/2$. Now suppose $n > m$. ...
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Find Lebesgue measurable sets such that the measure of intersection is product of measures.
First of all, $A_n = \emptyset$ or $(0,1)$ for all $n$ works but I guess you want a non trivial example.
The equality $m(A_n \cap A_k) = m(A_n)m(A_k)$ means, from the probabilitic point of view, that ...
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What is the Lebesgue measure of Luzin's non-Borel set of reals?
There exists a result for continued fractions similar of "normal" numbers: for almost every $x\in [0,1]$, the number $k$ occurs in the continued fraction of $k$ (in the limit) with ...
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What is the Lebesgue measure of Luzin's non-Borel set of reals?
I now believe justt's conjecture that the Lebesgue measure of $A$ is one, and I believe I have a proof. It uses some similar ideas to justt's proof outline (in particular, Khinchin's constant), but in ...
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Understanding proof of "if $f$ is non-negative, then it is the supremum of simple functions."
For your first question, yes that is true, but I am not sure why that would be helpful in proving $\int_B h \, d\mu \le \sup\{\cdots\}$.
For your second question: This follows from the definition of ...
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Optimal Transport Theory: Find a sequence of functions that converge weakly to 1/2
For (a), under the $\pi-\lambda$ Theorem, it suffices to prove $\mathcal{L}_{[0,1]}(T_n^{-1}([a,b]))=\mathcal{L}_{[0,1]}([a,b])$ for all intervals $[a,b]\subset[0,1]$, $n\in \mathbb{N}$ (here $\...
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Brezis' Proposition 8.4: an inequality that holds for a.e. $x,y \in \mathbb R$
Your proof is OK.
Maybe an easier proof is to show that $$A:=\{x\in\mathbb R: x\in N \text{ or } x+h\in N\}$$ has measure zero, hence for $\lambda\text{-a.e. }x\in\mathbb R$ we have $x\in N^c$ and $x+...
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What does it mean that a random variable has "Lebesgue density"?
Lebesgue density will refer to a "density" with respect to the Lebesgue measure. If you don't know much measure theory, then just read this as: "$Y$ has a probability density" (a ...
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Pointwise a.e. approximation by a sequence of smooth functions with supremum bound
The trick here is to use bump functions. Specifically, we are using the fact that if $K\subseteq U$ with $K$ compact and $U$ open, and $c_1$ and $c_2$ are constants, then there is a smooth (i.e., in $...
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Is the following operator well-defined?
No, I do not think it is.
And if by your observation that $u$ being square integrable implies it
being finite almost everywhere you mean that it must be bounded
(i.e., $|u(x,t)|\leq k$ for some $k$) ...
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