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9

If you want, you can think about this integral geometrically. The function you described is equal almost everywhere to the following step function (or should I say, function of infinite repeating narrower staircases; intervals are not drawn to scale): As you can see in the picture, what we get for the total area under that step function of infinite ...


5

Consider $S_d$ to be the set of all irrational $x$ numbers of the form $0.0\ldots 0dp_1p_2\ldots$ as their first non-zero digit. Then we can partition $S_d = S_d^1 \cup S_d^2\cup S_d^3\cup ...$ where the upper index indicates at which decimal point the first digit occurs. It is clear that $m(S_d^1) = 0.1$ (it is comprised of the irrationls in the interval ...


5

Hint: $$ \max(a,b)=\frac{a+b+|a-b|}{2},\quad a,b\in\mathbb{R}. $$


5

Just compute the integrals :). What is meant by this: For fixed $y$, we have \begin{align*} \int_0^1 \frac{x^2-y^2}{(x^2 + y^2)^2}\, dx &= \left[ -\frac{x}{x^2 + y^2}\right]_{x=0}^1\\ &= -\frac{1}{1+y^2} \end{align*} Hence \begin{align*} \int_0^1\int_0^1 \frac{x^2-y^2}{(x^2 + y^2)^2}\, dx\, dy &= -\int_0^1 \frac{1}{1+y^2}\, dy\\ &= ...


4

No, this isn't true. As an example: $$f(x)=\begin{cases}1 &\text{if } x\in (0,1/2] \\ 0&\text{else}\end{cases}$$ and $$g(x)=\begin{cases}1 &\text{if } x\in (1/2,1) \\ 0&\text{else}\end{cases}$$


4

That is correct, if the real numbers are a countable union of countable sets, then measure theory as we know it goes out the window. The same can be said if the real numbers are a countable union of countable union of countable sets, and so on. Essentially for the same reason. To your question, the Lebesgue measure is the completion of the Borel measure. So ...


4

Let $E_n = \{x: f_n(x)\neq g_n(x) \}$, so $E_n$ is a null set for all $n$. Let $E = \bigcup_n E_n$, so $E$ is also a null set (since countable unions of null sets are null sets). Define $f=\sup_nf_n$ and $g=\sup_n g_n$. Then $f(x)=g(x)$ if $x \notin E$ (because $f_n(x)=g_n(x)$ for all $n$ if $x \notin E$), so it follows that $\{ x: f(x) \neq g(x) \} ...


4

The linear map $(x, y) \mapsto (x + y, x - y)$ sends the graph of $f$ onto your set. The claim follows [Why?].


4

Consider using Holder's inequality with conjugate exponents $p$ and $p/(p-1)$: $$\left|\int_{A_n} f\, d\mu\right| \le \int_X \chi_{A_n}|f|\, d\mu \le \|\chi_{A_n}\|_{p/(p-1)} \|f\|_p = \mu(A_n)^{(p-1)/p} \|f\|_p.$$ Since $\mu(A_n) \to 0$, the result follows.


3

It looks right to me. You may notice that there is an even faster way to evaluate the integral: since the Canton set is symmetric with respect to $x=\frac{1}{2}$, $$ I=\int x\,d\mu_C = \frac{1}{2}\left(\int x\,d\mu_C+\int (1-x)\,d\mu_C\right)=\frac{1}{2}\int 1\,d\mu_C = \frac{1}{2}.$$ The Hausdorff dimension does not matter, we are simply putting a uniform ...


3

You're almost right. This is true for any measure which gives $0$ to singletons. It could be the Lebesgue measure, or some other regular Borel measure, or a measure giving $0$ to countable sets and $\infty$ to uncountable sets. But of course, there can be measure which give to some singleton, or some countable set a positive measure. For example the ...


3

The function $f$ is measurable if and only if there exists a sequence of step functions that converge to $f$ almost everywhere. Your $u$ is integrable, hence measurable. Therefore you have that sequence of step functions with support in $(0,1)$ which converges to $u$ a.e. Note that the same sequence converges a.e. to $\tilde u$ on $(0,\infty)$, hence ...


3

Lets consider $\Omega=\{1,2,3,4\}$ $\sigma(\{1\})$ is the smallest $\sigma$-algebra which contains $1$. So we must take any other elements of $P(\Omega)$ such that the conditions for being a $\sigma$-algebra are fulfilled. It does clearly contains $1$. Also $1^C=\{2,3,4\}$. And $\Omega,\emptyset$. So we have ...


3

I'm not sure you'll be able to find a text with solved exercises. My personal favorite is Folland's Real Analysis.


3

Counterexample: Consider $(0,1)$ with Lebesgue measure and the characteristic functions $\chi_{(0,\frac{1}{2})}$ and $\chi_{(\frac{1}{2},1)}$. Then $\chi_{(0,\frac{1}{2})}\chi_{(\frac{1}{2},1)} = 0$ but the product of their integrals is $\frac{1}{4}$.


2

The Dynkin-system is poor for the purpose of measure theory. In measure theory (probability theory) calculating the measure of a sets over which a sequence of functions converges is important. Let, for instance, $\{f_n\}$ is a sequence of measurable functions. Then te set $$\{x: \lim_{n \rightarrow \infty}f_n(x)=f(x)\}$$ cannot be expressed in general in a ...


2

In general, the claim does not hold true as the following counterexample shows: Consider $((0,1],\mathcal{B}(0,1])$ endowed with the Lebesgue measure. For $a_n := \frac{1}{n^2}$ we define $$g_n(\omega) := \begin{cases} \frac{1}{n}, & \omega \notin [a_n,a_{n-1}), \\ \frac{1}{\omega}, & \omega \in [a_n,a_{n-1}), \end{cases} \qquad \omega \in (0,1].$$ ...


2

$\dim F$ is not defined as a value for which the Hausdorff dimension equals 0. It's defined as the infimum of a set of such values, which does not mean it has to be a member of the set itself.


2

An example. $F = [0,7]$ with its usual metric. Then: $\mathcal H^s(F) = 0 $ for $s>1$, $\mathcal H^1(F) = 7$, $\mathcal H^s(F) = \infty$ for $0<s<1$. Thus, according to your definition, $\dim F = \inf\;(1,\infty) = 1$.


2

No, not even with large cardinals. Take any non-measurable set $A\subseteq\Bbb R$, and consider $\{0\}\times A$ as a subset of $\Bbb R^2$. As a subset of $\Bbb R^2$ it is a subset of $\{0\}\times\Bbb R$ which is a Borel set which is null. So $\{0\}\times A$ is Lebesgue measurable. But what is the projection of $\{0\}\times A$ onto $\Bbb R$?


2

As usual, the proof goes wrong when it is written that something is "clear": "Then it is clear that the family of all such intervals contain $[0,1]$ ..." Not only is this not clear, it's not true! (Try to prove it, you will see where the issue arises.)


2

Finite additivity and countable sub-additivity is equivalent to countable additivity. The proof is below. Let $\mu$ be a measure. It is clear that if $\mu$ is countably additive then it is finitely additive and countably sub-additive. Assume that $\mu$ is countably sub-additive and finitely additive. Consider a collection $\{A_n \}_{n=1}^\infty$ of ...


2

Any subset of $\mathbb R$ is a subset of a Borel set (namely $\mathbb R$). So any non-Lebesgue measurable set of reals is a counterexample.


2

Say we are doing this on the real line $\mathbb R$. Let $\mathcal G$ be the collection of all open sets. We are interested in $\sigma$-algebras $\mathcal F$ such that $\mathcal F \supseteq \mathcal G$. There may be many such $\sigma$-algebras. For example, the power set $\mathcal P$, consisting of all subsets of $\mathbb R$ is one. But that is the ...


2

Two standard arguments: (1) In order to show this equation holds for all $f,g \in L^2$, it suffices to prove it for all $f,g$ in some dense subset $V \subseteq L^2$. This is just a continuity argument (one should be a bit suspicious of the limit on the left, but it's not hard to deal with since $T$ is unitary). (2) Since both sides are bilinear in $(f,g)$, ...


2

Look at the inverse image of a generating measurable set. $$\bar{u}^{-1}(a,b)$$ If $a<0<b$ then because inverse images and unions commute, i.e. $f^{-1}(A\cup B)=f^{-1}(A)\cup f^{-1}(B)$ we get: $$\bar{u}^{-1}(a,b)=\bar{u}^{-1}(a,0)\cup \bar{u}^{-1}(\{0\})\cup\bar{u}^{-1}(0,b)$$ $$=u^{-1}(a,0)\cup (1,\infty)\cup u^{-1}(\{0\})\cup u^{-1}(0,b)$$ ...


2

By way of contradiction, assume that an integrable function $ f: \mathbb{R} \to [0,\infty) $ exists such that $$ \mu(E) = \mu(f^{\leftarrow}[E]) $$ for any Lebesgue-measurable subset $ E $ of $ [0,\infty) $. Then $$ \forall n \in \mathbb{N}: \quad \mu({f^{\leftarrow}}[n,n + 1)) = \mu([n,n + 1)) = 1. $$ Hence, \begin{align} ...


2

For non-negative measurable functions $f: \Bbb{R}^n \to [0, \infty]$ the Lebesgue measure of the set $$ \mathcal{G}(f) = \{ (x,y) \in \Bbb{R}^n \times [0, \infty) \mid 0 \leq y < f(x)\} \subset \Bbb{R}^{n+1} $$ is the integral of $f$, so \begin{equation} (*) \quad \int_{\Bbb{R}^n} f \, dm = m_{n+1}(\mathcal{G}(f))\,. \end{equation} It turns out that if we ...


2

If you want to avoid the subsequence argument you can use Chebyshev's inequality. For any $t > 0$ and $n \in \mathbb N$ you have $$\mu(\{|f_n - f| > t/2 \}) \le \frac{2^p}{t^p} \|f_n - f\|_p$$ and $$\mu(\{|f_n - g| > t/2 \}) \le \frac{2^{p'}}{t^{p'}} \|f_n - g\|_{p'}$$ so that $$\mu(\{|f - g| > t\}) \le \frac{2^p}{t^p} \|f_n - f\|_p + ...


2

Not true at all. As a matter of fact, if $\|f\|_1\not= 0$, the integral diverges! For example, consider $f(x)=\chi_{(0,1)}$. Then $\|f(x)\|_1=1$ and $$\int_{-\infty}^{\infty} 1-\chi_{(0,1)}dx=2\int_1^{\infty}1dx$$



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