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2 votes
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The set of the points where a measurable function is continuous is measurable

The continuity set of a real-valued function is $G_{\delta}$ i.e., a countable intersection of open sets. Open sets are measurable. Countable intersections of measurable sets are measurable.
MathematicsStudent1122's user avatar
3 votes

Find a sequence of $Lˆ1$-summable functions with $|f_n(x)| \leq 1$ and some other properties

A simple example, which you may have had in mind with your example, are rectangles that are getting wider and wider: $$f_n(x) := χ_{(0,n]}(x) \cdot \frac{1}{n}.$$ The area is obviously always 1 ...
Noctis's user avatar
  • 212
1 vote
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Find a sequence of $Lˆ1$-summable functions with $|f_n(x)| \leq 1$ and some other properties

When you are faced with these problems, always look for easy solutions first. The easiest example of integrable functions are $\frac{1}{x^n}$, however there are two issues: They are not integrable ...
Zima's user avatar
  • 3,392
0 votes

Lebesgue outer measure in $\mathbb{R}^2$ in terms of a grid of $h$-squares

No, it does not hold. For example, consider the set $$D=\big\{(x,y)\in\mathbb{Q}\mid0\leq x\leq 1,\ 0\leq y\leq 1\big\}.$$ $\lambda^\ast(D)=0$, but $A(D,h)=1$ for all $h>0$, so $\lim_{h\to0^+}A(D,h)...
ashpool's user avatar
  • 7,006
2 votes
Accepted

find measure of set $E=\cap_{n=1}^{\infty}\cup_{k=n}^{\infty} E_k.$ where $E_k$ has measure $\frac{1}{k^2}$.

One way to prove this is the following. We have $$\mu\left(\bigcup_{k=n}^\infty E_k\right)\leq \sum_{k=n}^\infty \mu(E_k)=\sum_{k=n}^\infty \frac1{k^2},$$ so for all natural $n$, $$\mu(E) \leq \mu \...
Jan's user avatar
  • 432
5 votes

Let $f\colon (a,b)\to \mathbb{R}$ be nondecreasing and continuous. If $E=\{x\in (a,b)\mid f'(x)\text{ exists and } f'(x)=0\}$, then $\lambda(f(E))=0$

The following proof is a more explicit version of the one already mentioned above. If anything remains unclear, I’ll be happy to provide clarification. Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be an ...
Jan's user avatar
  • 432
1 vote
Accepted

$\phi$ is continuous and odd, $f$ is Lebesgue integrable, we define $T_nf = \int_{R} \phi(nx)f(x)$. Show that: $lim_{n \to \infty} T_nf = 0$

This can be solved by direct application of Fejer's lemma: Theorem (Féjer): Suppose $\phi$ is a bounded measurable $T$ periodic function ($T>0$). For any $f\in\mathcal{L}_1(\mathbb{R})$ and ...
Mittens's user avatar
  • 40.6k
0 votes

A measurable bijection between the interval and the square

Maybe some idea: let's consider the decimal representation and the following function: $[0,1)\ni0. w x y z \ldots \mapsto (0. w y \ldots,0. x z \ldots) \in [0,1)^2$ This is clearly bijetive but why is ...
Riel Blakcori's user avatar
2 votes

Show $0<\epsilon<1$ there exists some $\delta_{p,\epsilon}>0$ such that $m(\{x\in X:|f(x)|>\epsilon\})\geq\delta_{p,\epsilon}$ for each $f\in E_p$.

Notice that for $f\in E_p$ and $0<\varepsilon<1$, \begin{align}1&=\int_X|f|=\int_{\{|f|\leq\varepsilon\}}|f|+\int_{\{|f|>\varepsilon\}}|f|\\ &\leq \varepsilon +\|f\|_p\big(\mu(|f|>\...
Mittens's user avatar
  • 40.6k
0 votes

Rudin $8.18$ theorem.

I don't understand how do we get that $\frac{Ap}{1-c} \int_0^{\infty} t^{p-2} dt \int_{E_i} f dm$ is equal to $ \frac{Ap}{1-c} \int_{R^k} f dm \int_0^{f/c} t^{p-2} dt$ It is by application of the ...
kobe's user avatar
  • 42.3k
1 vote
Accepted

when does $m(A/E)=m(A)-m(E)$?

(I'll assume with m(E) you mean $\mathscr{L}^n(E)$ the Lebesgue-measure and E $\subseteq$ $R^n$) You're right. If we have m(E)= $\infty$ we can't just say m(A\E) = m(A) - m(E) < $\epsilon$. We have ...
Rodrigo Calixto's user avatar
2 votes

Function $f:[0,1] \rightarrow \Bbb R$ such that $\forall c \in \Bbb R: f^{-1}(c)$ Lebesgue Measurable, but $f$ not Lebesgue Measurable.

As noted by Simon Pitte, your function is a correct counterexample, but as Anne Bauval noted, proving that $P$ is in bijection with $[0,1/2)$ is tricky (even though not too difficult). I think this ...
X-Rui's user avatar
  • 1,586
2 votes

Function $f:[0,1] \rightarrow \Bbb R$ such that $\forall c \in \Bbb R: f^{-1}(c)$ Lebesgue Measurable, but $f$ not Lebesgue Measurable.

"I know that for two uncountable sets $A,B$ there must exist a bijection between the two by transitivity, since for both there exists a bijection to $2^ω$ ." Not really: first, you certainly ...
Anne Bauval's user avatar
  • 39.5k
1 vote
Accepted

Function $f:[0,1] \rightarrow \Bbb R$ such that $\forall c \in \Bbb R: f^{-1}(c)$ Lebesgue Measurable, but $f$ not Lebesgue Measurable.

Your function is well-defined: set-theoretically, a function $f:A\rightarrow B$ is formally defined as a triplet $(A,B,G)$, where $A$ and $B$ are sets, and $G$ (the graph of the function) is a subset ...
Simon Pitte's user avatar
1 vote
Accepted

Continuous function which satisfies the Luzin N property, but which does not satisfy the Banach S property

For an $\epsilon>0$ and finite unions of half-open intervals (FOHOI for short) $J=\cup_{n=1}^N [a_n, b_n)$ and $J'=\cup_{k=1}^K [c_k, d_k)$, where $a_1<b_1<a_2<\cdots <a_N<b_N$ and $...
Jonathan Hole's user avatar
0 votes

How to show lim sup is $0$ for $f \in L^1(R)$

Let $b \in \mathbb{R}$ and let $t > 0$. Given an interval $I$ of length $t$ containing $b$, we have $$ \begin{align} \frac1{|I|} \int_I |f - f_I| &\leq \frac1{|I|} \int_I |f - f(b)| + \frac1{|I|...
Polygon's user avatar
  • 1,929
1 vote
Accepted

if $f_n \uparrow f$ prove $\mu(f_n\geq t)\rightarrow \mu(f\geq t)$

What you are trying to show is that $\int_{X}\mathbf{1}_{f_{n}\geq t}\to \int_{X}\mathbf{1}_{f\geq t}$. Probabilistically speaking, this ammounts to showing that for an increasing sequence $X_{n}$ of ...
Mr.Gandalf Sauron's user avatar
2 votes

if $f_n \uparrow f$ prove $\mu(f_n\geq t)\rightarrow \mu(f\geq t)$

$\mu(f_n >t) \le \mu(f > t)$ and $\mu(f > t) \le \mu(\cup_k \{f_k > t\}) = \lim_k \mu(f_k >t)$
Aristodog's user avatar
  • 373
2 votes

$L^\infty(\Omega)$ is dense in $L^{p,\infty}(\Omega)$ if $\Omega$ is compact

Actually, the closure of $L^\infty$ in $L^{p,\infty}$ is $$ L^{p,\infty}_0:=\left\{ f:\lim_{t\to\infty}t^p\mu\{x:\lvert f(x)\rvert>t\}=0 \right\}. $$ Indeed, if $f$ belongs to the closure of $L^\...
Davide Giraudo's user avatar
1 vote

$L^\infty(\Omega)$ is dense in $L^{p,\infty}(\Omega)$ if $\Omega$ is compact

This density is not true. Let $\Omega=(0,+1)$. Let $p\in (0,\infty)$ and define $$ f(x) = x^{-\frac1p}. $$ Then $$ \mu(\{x:\ |f(x)| > t\} ) = t^{-p} $$ for $t\ge1$, so that $\|f\|_{L^{p,\infty}} = ...
daw's user avatar
  • 50.3k

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