6

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 $f$ is non-differentiable on $E$, note that $(f(x+h) - f(x))/h \ge 1 + n$ if $h > 0$ and $(x, x+h) \subseteq U_n$.


6

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 $\mu(f^{-1}(I_1))\ge\mu(f^{-1}(I_0))/2$. Keep splitting the interval in half at the midpoint and choosing the half whose inverse image has at least half the ...


6

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 exists $N$ such that $A=\mu(E_N)>0$. $\chi(E_N) f^k\leq f^k$ implies that $\int\chi(E_n)f^k\geq A(1+{1\over N})^k\leq M$ for every integer $k$. ...


5

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 $M$-measures to $M_0$. These are both measures on $M_0$; and, since $\nu\ll\mu$, we also have $\nu_0\ll\mu_0$.


5

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 reals; this was originally studied by Addison, and has since then been a fundamental pillar of descriptive set theory and computability theory. This is a bit ...


5

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 nonmeasurable set. Then $N\times\{0\}$ is a Lebesgue measurable subset of $\mathbb{R}^2$ because the Lebesgue measure is complete, $N\times\{0\}\subseteq[0,1]\times\{...


5

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 form $(b,\infty)$ is measurable. So, suppose that $b \in \mathbb{R}$. If $b \in \mathbb{R}$, then we have \begin{align} g^{-1}((b,\infty)) &= \left\{ x : g(...


5

An uninteresting example: $f=0$ and $g=\chi_{E}$, where $E$ is any non-measurable set, then $f\circ g=0$.


5

I think that $\limsup f_n$ and $\liminf f_n$ are measurable and your set is the inverse image of $\{ 0 \}$ by the measurable function $g=\limsup f_n - \liminf f_n$.


5

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 Borel sets (generated by open intervals). This means that the preimage of an open interval like $(a,b)$ is measurable, but the preimage of a Lebesgue ...


5

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 $f_n: X\rightarrow Y$, which converges everywhere point-wise. Then we ruin $Y$. Let $f_1$ be constant $0$ on $[0,1/2[$ and constant $1/2$ on $[1/2,1]$. Let $f_2$ be constant $0$ on $[...


5

It doesn't follow from AC or from its negation. In fact it is refuted by AC, as you pointed out, e.g. the Vitali set. It doesn't follow from the negation, because a section of $\mathbb{R}\to\mathbb{R/Q}$ is enough to perform the Vitali construction and certainly AC doesn't follow from such a section existing. What is true is that under certain large ...


5

Hint: $g$ is the pointwise limit of a sequence of Riemann sums $$g_n(x) = \frac{1}{n} \sum_{k=1}^n f(x, k/n) $$


4

One problem with the "general definition" is that a continuous function might not be measurable. There is, for example, a continuous injective function $f$ from $\mathbb R$ to $\mathbb R$ that is a homeomorphism from a closed set $S$ of positive measure to the Cantor set $E$ (which has measure $0$). For any $V \subseteq S$, $V = f^{-1}(f(V))$, but $f(V) \...


4

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: $$f(x,y) = \begin{cases} 0 & \text{if } x \notin A \text{ or if } x \neq y , \\ 1 & \text{otherwise}. \end{cases} $$ One can easily check that $f$ is separately ...


4

The accepted answer is slightly incorrect, since there might not exist an $\epsilon$ such that $\{\omega \in \mathcal{X}: g(\omega)-f(\omega) > 0\} = \{\omega \in \mathcal{X}: g(\omega)-f(\omega) > \epsilon\}$. For example, let $\mathcal{X}=\mathbb{N}$, $f \equiv 0$ and $g(n)=n^{-1}$. However, the argument can be fixed. Let $A_{m}=\{g-f > m^{-1}\}$....


4

$\sigma$-algebras can contain much more than just a countable collection of sets - even for a countable basis, the collection of all countable unions and intersections is uncountable. For example, take the basis to be $\mathbb{N}$, then $\mathcal{P}\left(\mathbb{N}\right)\subset\sigma\left(\mathbb{N}\right)$, and $\left|\mathcal{P}\left(\mathbb{N}\right)\...


4

"Measurable" is the minimal requirement of a function to even think about its Lebesgue integral. You might have seen Wikipedia's page, and in particular the following picture; The lowest picture describes Lebesgue's integration. There, the base of every rectangle is a set of the form $$\tag{1} \{x\ :\ \lambda_1\le f(x)\le \lambda_2\},$$ and the integral ...


4

If we choose the product-$\sigma$-algebra on $\mathbb{R}^\mathbb{R}$, then any measurable set is the pre-image of a projection $\pi_I \colon \mathbb{R}^\mathbb{R} \rightarrow \mathbb{R}^I$, where $I$ is countable set. (This can be easily shown by using the definition of the product-$\sigma$-algebra. Any such projection is measurable. On the other hand, the ...


3

Very roughly, the idea is to approximate the uniform measure as a normalized sum of ``delta" measures. Here is one construction that is very easy to understand (I hope). For each $n$, let $A_n =\cup_{k=0}^{2^{n}-1} [\frac{k}{2^n},\frac{k}{2^n} + \frac{1}{2^{2n}}]$. Set $g_n = 2^n {\bf 1}_{A_n}$. Then $g_n \ge 0$, $\int g_n =2^n \times 2^{n} 2^{-2n}=1$, ...


3

Consider lemma Peter : if $g:\mathbb{R}\rightarrow \mathbb{R}^+$ is measurable, then $\mu \mapsto \mathbb{E}g(\mu)=\int g d\mu$ will be $D(\mathbb{R})\rightarrow \overline{\mathbb{R}}^+$ measurable. with the help of this lemma, you may prove your desired result using the hints/procedure written below Convince yourself that your result will follow from ...


3

Here is a version of the functional monotone class theorem found in Pollard's "A User Guide to Measure Theoretic Probability" Theorem. Let $\mathcal{H}^+$ be a $\lambda$-cone of bounded, nonnegative functions, and $\mathcal{G}$ be a subclass of $\mathcal{H}^+$ that is stable under the formation of pointwise products of pairs of functions. Then $\mathcal{H}...


3

I will outline the solution for the case $n=1$ and $p=2$, and if you can not generalize it to $p\ge1$ and $n\ge 1$, ask in comments, we will dive into the murky details. Without further ado: Let $K$ be a non-trivial compact set, $\phi$ be a test function with support in $K$ and $u'\in L^2(K)\cap D'(\Bbb R)$. We can say that by Cauchy-Bunyakovskiy-Schwarz ...


3

The interval $(0,1/4]$ contains a nonmeasurable subset (meaning a non Lebesgue measurable subset). The union of this set with $(1/4,1/2]$ is then a nonmeasurable subset $E$ of $(0,1/2]$ that has cardinality $c.$ Let $K_1,K_2$ be two disjoint Cantor sets, each a subset of $(0,1),$ having measure $0.$ Each of these has cardinality $c.$ Note that $(0,1)\...


3

Hint: Take $$f_n:= 2^{n}\sum_{k=0}^{2^n-1} 1_{[k2^{-n},k2^{-n}+4^{-n}]}$$ and consider subintervals of $[0,1]$ whose endpoints are dyadic (why does this suffice?). Also notice that the measure of the set of points where $f$ is nonzero equals $2^{-n}$.


3

Let $X_n = f^{-1} (-n, n)$. Then note that $X_n \subset X_{n + 1}$ for every $n \in \mathbb{N}$. Also $\bigcup X_n = X$. By continuity of measure, $\mu (X_n) \to \mu (X)$. As $\mu (X) < + \infty$, for every $\varepsilon > 0$, there exists an $N \in \mathbb{N}$ such that whenever $n \ge N$, $\mu (X) - \mu (X_n) < \varepsilon$. Therefore $\mu (X_N^...


3

Given a set $X$, the indicator function on $X$ is denoted $1_X$ or $1[X]$, and it is the function into $\{0,1\}$ given by $1_X(x) = 1$ if $x \in X$, and $1_X(x) = 0$ otherwise. The $1$ is meant to denote "this function takes the value $1$ if its argument is in…". The domain of the function is determined by context; in the instance you gave, the domain is $\...


3

$\mathcal{X}_E^{-1}(\{0\}) = E^C$, which is perfectly measurable. Take any $A \in \mathcal{B}(\mathbb{R})$: $\mathcal{X}_E^{-1}(A) =\begin{cases} X, & 0,1 \in A \\ E, & 1 \in A, 0 \notin A \\ E^C, & 1 \notin A, 0 \in A \\ \emptyset, & o.w. \end{cases}$ all those sets are measurable, since $E$ is measurable.


3

Suppose $d>1$. $f(x_1,\dots,x_d) := x_1$ is a measurable map $\mathbb R^d\to\mathbb R$. Furthermore, the set $$E := \{x\in \mathbb R^d \mid x_2 = x_3 = \dots =x_d = 0 \}$$ is null, but $f(E) = \mathbb R$.


3

It is not locally convex. Example. Let's take $L^0[0,1]$, equivalence classes mod null sets of measurable functions, with base for the neighborhoods of zero given by $$ V_\varepsilon = \big\{f \in L^0 : \lambda\{\,|f(x)| > \varepsilon\} < \varepsilon\big\} $$ for $\epsilon > 0$. I wrote $\lambda$ for Lebesgue measure. LINK We claim that the ...


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