Problem 7 Let $f \in$ \(\mathcal{L}^{1}\... [FREE SOLUTION] (2024)

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Chapter 13: Problem 7

Let $f \in$ $\mathcal{L}^{1}\left(\mathbb{R}^{d}\right), \mu=f \lambda^{d}$and let $C \subset \mathbb{R}^{d}$ be open, convex and bounded with $0 \in C.$ Show that $$ \lim _{r \downarrow 0} \frac{\mu(x+r C)}{r^{d} \lambda^{d}(C)}=f(x) \quad\text { for } \lambda^{d} \text {-almost all } x \in \mathbb{R}^{d}. $$ For the case $d=1$, conclude the fundamental theorem of calculus: $$ \frac{d}{d x} \int_{[0, x]} f d \lambda=f(x) \quad \text { for } \lambda \text{-almost all } x \in \mathbb{R}. $$ Hint: Use Exercise 13.1.6 with $\mu_{q}(d x)=(f(x)-q)^{+} \lambda^{d}(d x)$for $q \in \mathbb{Q}$, as well as the inequality $$ \frac{\mu(x+r C)}{r^{d} \lambda^{d}(C)} \leq q+\frac{\mu_{q}(x+r C)}{r^{d}\lambda^{d}(C)}. $$

Short Answer

Expert verified

The lemma is proved by applying Exercise 13.1.6 to a measure $\mu_q$ constructed based on $f$, and using an inequality given in the hint to prove that $\lim_{r \downarrow 0} \mu(x+r C)/r^{d} = f(x)$ for almost all $x$. The fundamental theorem of calculus in measure theory for $d=1$ is proven by taking $C=(0, 1)$ and using the definitions of measures $\mu$ and $\mu_q$.

Step by step solution

Preliminary Definitions

First off, let's note the definitions and notations in the problem. Here, $f \in \mathcal{L}^{1}(\mathbb{R}^{d})$ means that $f$ is a function in the space of Lebesgue integrable functions on $\mathbb{R}^{d}$, the $d$-dimensional real numbers. $\mu=f \lambda^{d}$ means that $\mu$ is a measure defined by multiplying the Lebesgue measure $\lambda^{d}$ by $f$. $C$ is a convex set containing the origin, and $r$ is a scalar that tends to zero.

Define measures $\mu_{q}$

According to the hint, we define a measure $\mu_{q}$ for $q \in \mathbb{Q}$ (the set of rational numbers) as $\mu_{q}(d x)=(f(x)-q)^{+} \lambda^{d}(d x)$, where (f(x)-q)^{+} denotes the positive part of $f(x)-q$, i.e. $(f(x)-q)^{+}=\max(f(x)-q,0)$.

Apply Exercise 13.1.6

The next step is to apply Exercise 13.1.6 to the measures $\mu_{q}$. Exercise 13.1.6 states that if $C \subset \mathbb{R}^{d}$ is open, convex, and bounded with 0 \in C, then $\lim _{r \downarrow 0} \mu_{q}(x+r C) / r^{d} =0$. Using this, we can derive that For any fixed ration number $q$, $\,\lim_{r\downarrow 0}\, \mu_{q}(x+ rC)/r^{d} = 0$.

Use the hint's inequality

Now we use the given inequality: $\frac{\mu(x+r C)}{r^{d} \lambda^{d}(C)} \leq q+\frac{\mu_{q}(x+r C)}{r^{d}\lambda^{d}(C)}$. Combined with the previous step, we get $\lim_{r \downarrow 0}\,\mu(x+ rC)/r^{d} \leq q$ for any rational number $q$. Since $q$ was arbitrary, this implies that $\lim_{r\downarrow 0}\, \mu(x+ rC)/r^{d}$ is at most $f(x)$.

Obtain inequality in the opposite direction

To finish off the proof for the first part, we need to find an inequality in the opposite direction. For this, note that $\frac{\mu(x+ rC)}{r^{d}} \geq (f(x) - q)\frac{\lambda^{d}(rC)}{r^d} $ because $f(x)\lambda^{d}(dx)$ is the measure of the set where $f-q > 0$. By taking the limit as $r \downarrow 0$, and using the fact that $\lim_{r \downarrow 0} \lambda^{d}(rC)/r^{d} = \lambda^{d}(C)$, we get $\lim_{r \downarrow 0} \mu(x+ rC) / r^{d} \geq f(x) - q$ for any rational $q$. Since $q$ can be taken as close to $f(x)$ as we want because $q$ is a rational number, we obtain that $\lim_{r \downarrow 0} \mu(x+ rC) / r^{d}$ is at least $f(x)$.

Apply the results to the case $d=1$

For the case $d=1$, we have $C=(0,1)$. By plugging in $C=(0,1)$, and deriving the integral from the definitions of measures $\mu$ and $\mu_{q}$, we can conclude that $\frac{d}{d x} \int_{[0, x]} f d \lambda=f(x)$ for $d$-almost all $x \in \mathbb{R}$, which is the fundamental theorem of calculus in measure theory.

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$Problem 7 Let $f \in$ \(\mathcal{L}^{1}\... [FREE SOLUTION] (3)$

Most popular questions from this chapter

Let $X, X_{1}, X_{2}, \ldots$ and $Y_{1}, Y_{2}, \ldots$ be real randomvariables. Assume $\mathbf{P}_{Y_{n}}=\mathcal{N}_{0,1 / n}$ for all $n \in\mathbb{N}$. Show that $X_{n} \stackrel{\mathcal{D}}{\longrightarrow} X$ ifand only if $X_{n}+Y_{n} \stackrel{\mathcal{D}}{\longrightarrow} X .$Let $x=\left(x^{1}, \ldots, x^{d}\right) \in \mathbb{R}^{d}$ and $y=$$\left(y^{1}, \ldots, y^{d}\right) \in \mathbb{R}^{d} .$ Recall the notation$x \leq y$ if $x^{i} \leq y^{i}$ for all $i=1, \ldots, d$. A$\operatorname{map} F: \mathbb{R}^{d} \rightarrow \mathbb{R}$ is calledmonotone increasing if $F(x) \leq F(y)$ whenever $x \leq y$. $F$ is calledright continuous if $F(x)=\lim _{n \rightarrow \infty} F\left(x_{n}\right)$for all $x \in \mathbb{R}^{d}$ and every sequence $\left(x_{n}\right)_{n \in\mathbb{N}}$ in $\mathbb{R}^{d}$ with $x_{1} \geq x_{2} \geq x_{3} \geq\ldots$ and $x=\lim _{n \rightarrow \infty} x_{n}$. By $V_{d}$ denote the setof monotone increasing, bounded right continuous functions on$\mathbb{R}^{d}$. (i) Show the validity of Helly's theorem with $V$ replaced by $V_{d}$. (ii) Conclude that Prohorov's theorem holds for $E=\mathbb{R}^{d}$.For two probability distribution functions $F$ and $G$ on $\mathbb{R}$, definethe Lévy distance by $d(F, G)=\inf \\{\varepsilon \geq 0: G(x-\varepsilon)-\varepsilon \leq F(x)\leq G(x+\varepsilon)+\varepsilon$ for all $x \in \mathbb{R}\\}$ Show the following: (i) $d$ is a metric on the set of distribution functions. (ii) $F_{n} \stackrel{n \rightarrow \infty}{=} F$ if and only if$d\left(F_{n}, F\right) \stackrel{n \rightarrow \infty}{\longrightarrow} 0$. (iii) For every $P \in \mathcal{M}_{1}(\mathbb{R})$, there is a sequence$\left(P_{n}\right)_{n \in \mathbb{N}}$ in $\mathcal{M}_{1}(\mathbb{R})$ suchthat each $P_{n}$ has finite support and such that $P_{n} \stackrel{n\rightarrow \infty}{\Longrightarrow} P$.Let $\Omega$ be a Polish space, let $\mu$ be a $\sigma$-finite measure on$(\Omega, \mathcal{B}(\Omega))$ and let $f: \Omega \rightarrow \mathbb{R}$ bea map. Show that the following two statements are equivalent: (i) There is a Borel measurable map $g: \Omega \rightarrow \mathbb{R}$ with$f=g \mu$-almost everywhere. (ii) For any $\varepsilon>0$, there is a compact set $K_{\varepsilon}$ with$\mu\left(\Omega \backslash K_{\varepsilon}\right)<\varepsilon$ such that therestricted function $\left.f\right|_{K_{\varepsilon}}$ is continuous.Let $E=\mathbb{R}$ and $\lambda$ be the Lebesgue measure on $\mathbb{R}$. For$n \in \mathbb{N}$, let $\mu_{n}=\left.\lambda\right|_{[-n, n]}$. Show that$\lambda=\operatorname{v}_{n \rightarrow \infty} \mu_{n}$ but that$\left(\mu_{n}\right)_{n \in \mathbb{N}}$ does not converge weakly.

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Problem 7 Let \(f \in\) \(\mathcal{L}^{1}\... [FREE SOLUTION] (2024)

Short Answer

Step by step solution

Preliminary Definitions

Define measures \(\mu_{q}\)

Apply Exercise 13.1.6

Use the hint's inequality

Obtain inequality in the opposite direction

Apply the results to the case \(d=1\)

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Decision Maths

Mechanics Maths

Probability and Statistics

Calculus

Pure Maths

Theoretical and Mathematical Physics

Study anywhere. Anytime. Across all devices.

Privacy Overview

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