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5

First note $f$ is onto: Otherwise the range of $f$ is a proper subinterval $[a,b],$ which implies $\lambda(f^{-1}([a,b])) = \lambda([0,1]) = 1 > b-a = \lambda([a,b]),$ contradiction. Suppose $f'(x) = 0$ for some $x\in [0,1].$ Then $f(B(x,r))\subset B(f(x),r/4)$ for small $r>0.$ ($B(y,r)$ denotes the open ball with center $y$ and radius $r$ relative to ...


5

The reason why this is a contradiction is because this argument can be applied to any starting set. Therefore we wind up concluding that every set has measure zero, which is boring. Properly stated, the theorem says that any translation and rotation invarient finitely additive nontrivial measure on $\mathbb{R}^3$ has non-measurable sets.


5

Hint: $$\frac{x}{e^x-1}=\sum_{n=1}^\infty xe^{-nx}$$


5

Let $f_n=\sum_{k=1}^ng_k$. Then the sequence $\{f_n\}$ is increasing because the $g_k$ are non-negative, hence by the monotone convergence theorem $$ 0\leq \int \sum_{k=1}^{\infty}g_k\;d\mu=\sum_{k=1}^{\infty}\int g_k\;d\mu\leq \sum_{k=1}^{\infty}\frac{1}{k^2}<\infty$$ Since $\sum_{k=1}^{\infty}g_k$ is a non-negative function with finite integral, it ...


4

Consider $\Omega = [-1, 1]$, $X(t) = t$, and set the codomain of $Y$ to $\{0, 1\}$, with $$ Y(t) = \begin{cases} 0 & t \le 0 \\ 1 & \text{else} \end{cases}. $$ Let $x = 0$ and $y = 0$ in your last paragraph. The first set is $\{0\}$ and the second is $[-1, 0]$, and their intersection is $\{0\}$. So the answer is "no" to that main question. It ...


3

Yes, this is true by a simple continuity argument. Note that for any fixed $x\in\mathbb{R}^2$, the map $p\mapsto \|x\|_p$ is a continuous function $[1,\infty]\to\mathbb{R}$ (it is obvious that this is continuous for $p<\infty$; continuity as $p\to \infty$ requires a little work but is not hard). So by the intermediate value theorem, if $\|x\|_1>1$ ...


3

If $[h \sim c]$ denotes the set $\{x\mid h(x) \sim c\}$ (where $\sim$ is any relation), notice that your set is: $$[f - g = 2] = [f - g \le 2] \cap [f - g \ge 2]$$ both of which are measurable since $f - g$ is measurable.


3

For 1, just note that it's a telescoping sum. That is: $$\sum_{n=1}^k (nx^{n-1}-(n+1)x^n) =(1x^{1-1}-(1+1)x^1)+\cdots+(kx^{k-1}-(k+1)x^k)=1-(k+1)x^k$$ Just take the limit as $k \to \infty$ and you get $1$ For 2, just do the integral, The antiderivative is just the power rule.


3

Hint: The sequence of partial sums $\Big(\sum_{n=1}^{N}g_{n}\Big)_{N=1}^{\infty}$ is monotonically increasing because each $g_{n}$ is non-negative.


3

You need to know (or to prove) that $\sum \frac 1 {n^2}$ is finite, with sum $S$ (you don't need to know what the sum is - that is a much harder problem than just knowing the sum is finite). Now if $f_M = \sum_1^M g_n$, the $f_M$ form a monotonically increasing sequence (why?) and the integral of $f_M$ is bounded by S, so the integral of the limit (that is, ...


3

Recheck what $\lambda \perp \mu$ means. It means that there are disjoint (measurable) sets $A,B$ such that $\lambda$ is concentrated on $A$ and $\mu$ is concentrated on $B$, i.e. $\lambda(E) = 0$ whenever $E \cap A = \varnothing$, and $\mu(E) = 0$ whenever $E\cap B = \varnothing$. If $\lambda = \mu$, we can write $$E = (E\cap A) \cup (E\setminus A)$$ to ...


3

$$\mu(F_k)\leq\sum_{j=k}^\infty\mu(E_k)<\frac{2^{-k}}{1-\frac{1}{2}}=2^{-k+1}=2^{-(k-1)}$$


3

No, it does not: If $E$ is (bounded and) non-measurable, then $f = 2\chi_{E} - 1$ is everywhere equal to $\pm1$, so $|f| \equiv 1$, but $f$ is non-measurable, hence not integrable.


3

Yes the reverse holds as well. This is really a definitional convention. We typically define "$f$ is integrable" to mean that both $\int_\Omega f^+ d\mu$ and $\int_\Omega f^- d\mu$ are finite (since, in the beginning, we define integration only for non-negative functions) then define $$\int_\Omega f d\mu = \int_\Omega f^+ d\mu - \int_\Omega f^- d\mu.$$ ...


2

Define $g_M := \max(\min(g,M),-M) ∈ L^1 ∩ L^\infty$ and apply your argument modified so that $s_k → \operatorname{sgn} g_M$ in $L^1$ to find that $‖g_M‖_1 = 0$. On the other hand, $|g_M| \uparrow |g| ∈ L^1$, so MCT says $∫|g_M| → ∫ |g|$, which means $∫|g| = 0$. (Actually since $g_M^2 \uparrow g^2$ as well you don't need to modify your $s_k$ but I'll just ...


2

Here's the worst possible way to keep straight which direction the inequality goes: In a typical programming language strings are sorts alphabetically, so that for example "cat" < "dog", since "c" < "d". Observe that $$\text{IL < LI.}$$"Integral(limit) < Limit(integral)". This is in my opinion extremely awful, because it has nothing to do with ...


2

Take $$S_n=\sum_{k=1}^n\left|f_k\right|$$ and apply the monotone convergence theorem for $\left\{S_n\right\}_{n=1}^\infty$.


2

This is called continuity of measure from above. It applies for finite measures, including probability measures. Here is how to start the proof. Write $P \left ( \bigcap_{n=1}^\infty A_n \right ) = P \left ( \left ( \bigcup_{n=1}^\infty A_n^c \right )^c \right ) = 1-P \left ( \bigcup_{n=1}^\infty A_n^c \right )$. Define $B_k=A_k^c \setminus ...


2

In the paragraph above Lemma 2.6, the authors explicitly state "[we write] $\mathcal L^\infty$ for the space of measurable bounded functions"


2

Hint: $$\sum \int g_n = \int \sum g_n$$ Why? Also, if $\int f < \infty$ where $f \ge 0$, then $f < \infty$ a.e. (this is well-known)


2

Note that $E\times E$ has only 4 elements. For any $x\in E$ we have $$\mu_1\otimes\mu_2(\{(x,x)\}) =\mu_1(\{x\})\mu_2(\{x\})= \mu_2\otimes\mu_1(\{(x,x)\})$$ So we need to inspect only the two other points: $(0,1)$ and $(1,0)$. Then we have $$p(\mu_2(E)-q) = \mu_1\otimes\mu_2(\{(0,1)\}) = \mu_2\otimes\mu_1(\{(0,1)\}) =q(\mu_1(E)-p) \tag{1}$$ and ...


2

Let: $$g_n(t) = \frac{f(t + \frac1{n}) - f(t)}{\frac1{n}}$$ Then $(g_n)$ is a sequence of measurable functions which converges pointwise to $f'$ a.e. Hence $f'$ is (Lebesgue) measurable.


2

Note that $X$ is a finite set. So given $x\in X$, the set $B_x= \{ S \in \mathcal{F} : x\in S\}$ is finite. Since $\mathcal{F}$ is a $\sigma$-algebra, we have $X\in B_x$ ($B_x$ is not empty) and $$A_x=\bigcap_{S\in B_x} S \in \mathcal{F}$$ Clearly $x\in A_x$, and $A_x$ is the smallest element in $\mathcal{F}$ containing $\{x\}$. Note that if $y\in X$ and ...


2

Let $A_n$ be the natural numbers with the first $n$ natural numbers thrown out. I.e. $A_n=\mathbb N\setminus\{1,2\cdots n \}$. Then each $\mu(A_n)=\infty$ and $\cap A_n=\phi$ so that $\mu(\cap A_n)=0$. Can you adapt this argument to your example?


2

Choose $x_k \in \Omega$ such that each $x_k$ is distinct. Let $A_n = \{x_k\}_{ k \ge n}$.


2

There is a set $A \subset \Omega$ that can be identified as $\mathbb{Z}$ (in the sense that there is a bijection $f: \mathbb{Z} \to A$). Let $A_n = \mathbb{Z} \setminus \left\{ 1 ,\dots, n \right\}$. This sequence satisfies the claim.


2

Try this, which I think is similar to your strategy. Let $C=\{x_1, x_2, \cdots \}$ be a countably infinite set of distinct points. Let $A_n:= \{x_i \in C : i\geq n\}$. So $A_1=C$, $A_2=\{x_2, x_3,\cdots \}$, etc. Then Clearly $$\bigcap_n A_n =\emptyset$$ and $A_n \downarrow$. However, for each $n$, $\mu(A_n)=\infty$, so $\mu(A_n)$ cannot tend to 0.


2

To be true that $A_i \downarrow A \implies \mu(A_i) \downarrow \mu(A)$, you should have that $\mu(A_1) < \infty$. Hint: construct an example where $\mu(A_1) = \infty$.


2

By the monotone convergence theorem, you always will have $$\int f^p = \lim_{k \rightarrow \infty} \int f_k^p$$ Taking $p$th roots (and using continuity of the function $x \rightarrow x^{1 \over p}$), one therefore has $$||f||_p = \lim_{k \rightarrow \infty} ||f_k||_p \tag 1$$ Since the $f_k$ increase to $f$, the limit in $(1)$ is an increasing limit. So one ...


2

The following argument should work under the additional assumption that $f'$ is integrable. I don't know if this is something that you need to assume or if it is a limitation of my proof. It is worth mentioning that I need integrability of the derivative for two reasons: first to apply Lebesgue differentiation theorem for $f'$, and second to get that $f$ is ...



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