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12

Have $\mu$ be a $\sigma$-finite measure on the $\sigma$-algebra $\mathscr{M}$. Have $\mu^*$ be the outer measure induced by $\mu$, $\mathscr{M}^*$ the $\mu^*$-measurable sets and $\overline{\mu}$ the measure $\mu^*$ restricted to $\mathscr{M}^*$. We want to prove $\overline{\mu}$ on $\mathscr{M}^*$ is the completion of the measure $\mu$ defined on ...


6

I will show that $\mathscr M^*=\overline{\mathscr M}$. From Theorem 1.9 and Carathéodory's theorem it follows, indeed, that $\mu^*|\mathscr M^*$ is the completion of $\mu$, since $\mu^*|\mathscr M^*$ is a complete measure on the $\sigma$-algebra $\mathscr M^*=\overline{\mathscr M}$ and it extends $\mu$ by Proposition 1.13. Consider a measure space ...


5

Let $E_1 = [0, 1/2]$; $E_2 = [1/2, 1]$ $E_3 = [0,1/4]$; $E_4 = [1/4,1/2]$; $E_5 = [1/2,3/4]$; $E_6 = [3/4,1]$ $E_7 = [0,1/8]$; $E_8 = [1/8,1/4]$; etc. Then $\mu(E_n) \rightarrow 0$, but for any $x \in [0,1]$, it's easy to see that $f_n(x) = f(x)\chi_{E_n^c}(x)$ differs from $f(x)$ for infinitely many $n$, so $f_n(x)$ does not converge pointwise to $f(x)$ ...


5

Consider $f(x)=x$ and $g(x)=1-x$ on $E=[0,1]$.


4

Why do you claim a contradiction in the last two lines? The hypothesis is that $$\int_E f = \int_E g$$ but it need not be the case that $$\int_{E_n} f = \int_{E_n} g$$ Consider, for example, $E = [0,2\pi]$, $f(x) = \cos(x)$, and $g(x) = \sin(x)$. Then $\int_E f = \int_E g = 0$ but, for example, $\int_0^\pi f \neq \int_0^\pi g$.


3

Use the Dominated convergence theorem. $ f_ n$ converges pointwise to $0$ and is bounded by $1/\sqrt x$ which is integrable.


3

Let $A = A_0 \cup A_1$, where $A_0 = \{x\in X : \lim f_n(x) < \infty\}$ and $A_1 = \{x\in X : \lim f_n(x) = \infty\}$. Then $x \in A_0$ if and only if for every $j\in \Bbb N$, there exists an $N \in \Bbb N$ such that $|f_m(x) - f_n(x)| < \frac{1}{j}$ for all $m, n \ge N$. Thus $$A_0 = \bigcap_{j = 1}^\infty \bigcup_{N = 1}^\infty \bigcap_{m = N}^\infty ...


3

Here is an important thing to keep in mind: If you want to show that the function $$ \Bbb{R} \to \Bbb{R}, x \mapsto e^{x^2 \cdot \sin(x)} - \sum_{n=0}^\infty \frac{x^n}{(n!)^2} $$ is continuous, you will (hopefully) not try to find for every $\varepsilon > 0$ some $\delta > 0$ such that ... Instead, you will rely on certain closure properties of ...


3

First note that the set $S$ is a subset of the interval $[-a, a]$, where $$ a = \sum_{n=1}^\infty e^{-n} = \frac{1}{e - 1} \approx 0.582 $$ The set is self-similar via two affine transformations, each scaling down by a factor of $e$ then shifting either right $(+)$ or left $(-)$ by $e^{-1}$. In symbols, the functions $f_\pm(x) = e^{-1}x \pm e^{-1}$ cover ...


3

The costant functions are elements of $L^{\infty}$ in lebesgue measure on $\mathbb{R}$, but they're not $L^2$, unless they're not $0$. It is true that a function $f\in L^{\infty}(\mu)$ is also in $L^2$ if the domain of the function has finite measure. Answering your question: if $f\in L^{\infty}$ and $g\in L^2$, then $$\|fg\|_2^2=\int ...


3

Let $f: [0,1] \to [0, \infty)$ be the characteristic function of the set of irrationals in $[0, 1]$. Assume there exists a nondecreasing sequence of step functions $\{s_n\}$ converging pointwise to $f$. Without loss of generality we can assume that $n \ge 0$. Since rationals are dense and $\{s_n\}$ is a nondecreasing sequence of step functions, $s_n(x)$ ...


3

A function is absolutely continuous if and only if it is an antiderivative of some integrable function (this is the approach via measure theory, which I assume is what you want: it connects to the property $\ m(A) < \delta \Rightarrow \int_{A} f < \epsilon$ that you mentioned). Given $g(x)=x|\sin(x^{-1/2})|$, you can find it's differentiable ...


3

It means that they satisfy $A_n \subset A_{n+1}$. The limit is not a limit in the $\epsilon$-$\delta$ sense. It just means $\cup_{n=1}^\infty A_n$. For example, take $A_n = [0,n]$, then $\cup_{n=1}^\infty A_n = [0, \infty)$. From a probability perspective, there is a real limit associated with these nested sets. Suppose $p$ is the probability measure. ...


3

For this problem we do not actually need $f \ge 0$. Recall that a sequence in a metric space $Y$ converges to a point $p$ in a metric space if and only if every subsequence has a subsequence converging to $p$. Here our metric space is $Y = \mathbb{R}$, our point is $p=0$, and our sequence will be the integrals of $\int \chi_{E_n}$, where $E_n$ sequence of ...


3

In my comment above, I noted $$\int_0^{1/2} f(t) \mathop{dt} - \int_{1/2}^1 f(t) \mathop{dt} \le \|f\|_\infty.$$ I have a hunch that the definition of $M$ should be $$\int_0^{1/2} f(t) \mathop{dt} - \int_{1/2}^1 f(t) \mathop{dt}=1/2$$ in order for part ii) to work. [Or more generally, the number in the definition of $M$ and the number in part ii) should be ...


2

Hint: $X_1 + \ldots + X_n < t$ iff there exist rational numbers $r_1, \ldots, r_n$ such that $r_1 + \ldots+ r_n < t$ and all $X_i \le r_i$.


2

This is the proof given in Stein's text. Let $f_n:= f \cdot \chi_{\{f \le n\}}$ (in other words, $f_n$ equals $f$ whenever $f$ is less than $K$, and is zero elsewhere). Then the $(f_n)_{n \ge 1}$ form an increasing sequence, so by the monotone convergence theorem, $$\lim_{n \to \infty} \int f_n \mathop{d\mu} = \int f \mathop{d\mu}.$$ Fix $\epsilon>0$. ...


2

For the first part, since $f$ is bounded: $$ |F(x)| = \left|\int_{\mathbb R} K(x-s)f(s)\,ds\right| = \left| \int_{\mathbb R} f(x-s)K(s)\,ds \right| \le M \int_\mathbb R |K(s)|\,ds $$ where $M = \sup_\mathbb{R} |f|$. (The absolute values are missing in @Ant's solution.) For the second part, \begin{align} \lim_{x \to \pm\infty} F(x) &= \lim_{x \to ...


2

Your argument for (b) is correct. You can use the same idea for (a); this time let $$A_N=\{x\in X:|f_n(x)|\le N\text{ for all }n\in\Bbb Z^+\}\;.$$


2

Assume that $X=\{x_j,j\in\mathbb N\}$ and that $\mu\{x_j\}>0$ for each $j$ and $f=0$, $f_n\geqslant 0$. We have to show that for each $j$, $f_n(x_j)\to 0$ as $n$ goes to infinity. Fix $j$ and $\varepsilon\gt 0$; for $n$ large enough, we have $\mu\{x,f_n(x)\geqslant\varepsilon\}\lt\mu\{x_j\}$, hence $f_n(x_j)\lt\varepsilon$ for these $n$. In general, ...


2

If we accept that $\frac{d}{dx}e^x=e^x$, then we can simply use implicit differentiation. $$\begin{align}y&=\ln x\\x&=e^y\\\frac{d}{dx}x&=\frac d {dx}e^y\\1&=\frac{dy}{dx}e^y\\\frac 1{e^y}&=\frac{dy}{dx}\\\frac{dy}{dx}&=\frac 1 x\end{align}$$


2

Suppose we want to find the antiderivative of $\cfrac 1x$ - it doesn't come in the pattern of derivatives of powers, because differentiating a constant always gives zero. When we look at the function we realise we will have a problem if our interval of integration includes zero, so we have to avoid that. And we see that this is an odd function, so if we have ...


2

Let $f:[0,1]\to[0,\infty)$ be given by $$f(x)=\begin{cases} 0 &; x\in\mathbb{Q} \\ 1 &; x\notin\mathbb{Q}\end{cases}$$ If $0\leq s\leq f$ is a step function, then $s$ is non-zero for at most finitely many points. Hence, any well-defined limit $\lim_{n\to\infty}s_n$ of step functions $0\leq s_n\leq f$ is non-zero for at most countably many points and ...


2

It depends. If the domain has finite measure, then it is true; in general $p > q \implies \mathcal L ^p(\Omega) \subset \mathcal L ^ q(\Omega) $ if $|\Omega| < \infty$ Note that with the lebesgue measure, it means that $\Omega$ is finite, but it need not to be that way; for example if you use a probability measure (or every measure such that ...


2

Let $E_{n,\varepsilon} = \{x : |f_n(x)-f(x)| > \varepsilon$. If $\mu(E_{n,\varepsilon}) > \delta$ then $$\int|f_n-f|^p\geqslant \int_{E_{n,\varepsilon}}|f_n-f|^p > \int_{E_{n,\varepsilon}} \varepsilon^p = \mu(E_{n,\varepsilon})\varepsilon^p>\varepsilon^p\delta, $$ and the desired inequality is obtained by raising both sides to the $\frac1p$ ...


2

Hint: Set $Y_k = B_{t_k + s} - B_{t_{k-1}+s}$ where $t_0 = 0$. Note that $X_k = Y_1 + \dots + Y_k$, so you can express $f(X_1, \dots, X_n)$ as a Borel function of $Y_1, \dots, Y_n$. Now observe that $\{\mathcal{F}_s, \sigma(Y_1), \dots, \sigma(Y_n)\}$ are mutually independent $\sigma$-fields. Use a Dynkin lemma argument to conclude that $\mathcal{F}_s$ ...


2

From $L^p$ to $\omega$ If $f\in L^p$ with $1<p\le \infty$, then by Hölder's inequality. $$\int_E |f| \le \|f\|_p \|\chi_E\|_q = \|f\|_p(\mu(E))^{1/q} $$ where as usual, $p^{-1}+q^{-1}=1$. So, in this case $f$ has the modulus of integrability of the form $\omega(\delta)=C\delta^{1-\frac{1}{p}}$. This bound is not perfectly tight for $p<\infty$: ...


2

We have $|f(x) + g(x)| \leq |f(x)| + |g(x)|$ for every $x$, by the triangle inequality. Now $|f(x)| \leq \|f\|_\infty$ almost everywhere, and similarly $|g(x)| \leq \|g\|_\infty$ almost everywhere. Therefore, $$|f(x) + g(x)| \leq \|f\|_\infty + \|g\|_\infty$$ almost everywhere. Since the right hand side is an almost-everywhere upper bound for $|f(x) + ...


1

Don't integrate over the whole line. Integrate over the interval $[-M,M]$: $$ \int_{[-M,M]} \sum_{k=1}^\infty \frac{a_k}{\sqrt{|x - \alpha_k|}} \, dx = \sum_{k=1}^\infty a_k \int_{[-M,M]} \frac{1}{\sqrt{|x - \alpha_k|}}\, dx.$$ The last integral does not exceed $4 \sqrt{M}$, so that the series has finite integral on $[-M,M]$, and is thus finite almost ...


1

In Minkowski's inequality, equality holds if and only if $f$ and $g$ are linearly dependent. If $f,g$ have norm $1$, the only way for them to be linearly dependent is if $f = \lambda g$ with $|\lambda| = 1$. Plugging this into your expression for $h$, we get $$h = \frac{1}{2}(f+g) = \frac{1}{2}(\lambda g+g) = \frac{\lambda+1}{2}g.$$ Then $$\|h\| = ...



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