43

A probability on $\mathbb{R}$, be it continuous or not, is given by its CDF $x \mapsto\mathbb{P}(X \leq x)$. A CDF is right-continuous, and the set of right-continuous functions has the cardinality of $\mathbb{R}$. To see this, you can for instance argue that the values of such a function are given by its values at the rational points, so it has at most the ...


34

You could utilize one of the well known ways to count the rational numbers, namely consider the integer lattice $\mathbb Z^2$ and the subset $\{(a,b)\mid a\geq 1\ \wedge\ b\geq 0\}$ as illustrated here: This corresponds to the positive rationals, namely $(a,b)\mapsto\frac ba$. It is a surjective covering and it is now simple to see how we might cover all ...


33

This is a really hard question; I think in general intuition for this sort of thing tends to come with experience, as you get used to the concepts. Having said that, I'll try to articulate the way that I think about it. I guess the way of viewing $\mathbb{Q}$ as a subset of $\mathbb{R}$ is a load of dots on a continuous line. Obviously these dots are very ...


28

This isn't a geometric answer, but you can get a lot of intuition for Lebesgue measure by thinking about it probabilistically. Specifically: The measure of a subset $S\subseteq [0,1)$ is the same as the probability that a randomly chosen point in $[0,1)$ will be an element of $S$. For example, the set $S = [0,1/4] \cup [5/8,3/4]$ has measure $3/8$ ...


15

No. We know that "Every set of reals has the Baire property" will also have the same consequence, and that ZF+"Every set of reals has the Baire property" is actually a weaker(!) theory than ZF+"Every set of reals is Lebesgue measurable", as shown by Shelah.


11

To expand on the AC/CH question, Raoul's argument does not depend on either of these, since you can give an explicit injection from real-valued sequences $x_1,x_2,\ldots$ to $\mathbb R$ (and there is an explicit bijection between $\mathbb Q$ and $\mathbb N$, so between $\mathbb R^{\mathbb Q}$ and $\mathbb R^{\mathbb N}$). To do this, write each value as an ...


11

Topologies and $\sigma$-algebras are designed with different objectives in mind. $\sigma$-algebras are designed to play nicely with measures, which are a generalized kind of volume measuring map. Topologies are designed to capture a notion of "closeness": when is a point $x$ close to a set $S$? If every open neighborhood of $x$ intersects $S$. When ...


10

$S$ only contains numbers with finite binary expansion. $\frac13$ is not there. The irrational numbers are not there. So the Lebesgue measure is $0$.


10

I know the existence of a non-Lebesgue measurable sets (i.e. Vitali set) having outer Lebesgue measure $0$ That's not true. Vitali sets have positive outer measure, and indeed that's how we prove they're not measurable (if they were, by the countable additivity of Lebesgue measure the interval $[0,1]$ would have to have infinite measure). It is indeed the ...


10

Every Riemann-integrable function is Lebesgue-integrable, so the only way in which the DCT could possibly fail for Riemann-integrable functions is in concluding that the limit function is Riemann-integrable. It is not too hard to cook up an example of a (dominated) sequence of Riemann-integrable functions whose limit is not Riemann-integrable: Let $\{r_1,r_2,...


8

Simply, since $$ S\subset\Bbb Q\;, $$ calling $\mu$ the Lebesgue measure, you get $$ 0\le\mu(S)\le\mu(\Bbb Q)=0\;. $$


8

This is a cute problem. Suppose both $$\int_{\mathbb R}|f(x)|dx<\infty\text{ and }\int_{\mathbb R}\frac{1}{|f(\frac 1 x)|}dx<\infty$$ hold. Note that this implies $f(x)\ne0$ for almost all $x$. Then a change of variables gives us $$\int_{\mathbb R}\frac{1}{|x|^2|f(x)|}dx<\infty$$ as well. Now by the Arithmetic-Geometric Mean inequality (that is, $\...


8

Let $X_1$ be any non-constant random variable, and let $X_2=-X_1$. For $f(x,y)=(x+y)^2$, we have $Ef(X_1,X_2)=E((X_1+X_2)^2)=0$ and $E(f(EX_1,X2))=E((EX_1+X_2)^2)=E((X_2-EX_2)^2)=Var(X_2)>0$.


8

I assert that the answer is undefined. Let's describe $(\Omega,\Sigma,\Pr)$, the probability space of $2^{\aleph_0}$ coins being tossed, one for each real number. The "Borel" sets in $\Sigma$ are the $\sigma$-algebra generated by sets of the form $\Pi_{r \in \mathbb R} F_r$ where finitely many $F_r$ are not $\emptyset$ or $\{H,T\}$, and the ...


7

Dominated convergence theorem of Riemann-integral is only valid when the limit function $f$ is assumed Riemann-integrable (the limit function will automatically be Lebesgue-integrable under the DCT-conditions).


7

No! Example: Let $\mathbb{P}(X>\alpha)=\alpha^{-1}(\log\alpha)^{-1+\delta}$ for all sufficiently large $\alpha$, with $\delta\in[0,1)$. Then $\int^\infty\mathbb{P}(X>\alpha)\,\mathrm{d}\alpha=+\infty$ and hence $\mathbb{E}X$ cannot be finite.


7

If $\mathcal F$ is complete, then it's true since $$f(x)=\limsup_{n\to \infty }f_n(x),\quad \mu-\text{a.e.}$$ Since $g_n(x):=\sup_{k\geq n}f_k(x)$ is decreasing, you get that $$\{f(x)\geq \alpha \}=\bigcap_{n\in\mathbb N}\{g_n(x)\geq \alpha \},$$ which is measurable because $g_n$ is measurable. Indeed, $$\{g_n(x)>\alpha \}=\bigcup_{k\geq n}\{f_k(x)>\...


7

If $f''$ is continuous (or more generally, if $f'$ is absolutely continuous) then $$ f(x+1) = f(x) + f'(x) + \int_x^{x+1} (x+1-t)f''(t) \, dt $$ from Taylor's theorem with the integral remainder. It follows that $f' \in {\cal L}^1$ if $$ g(x) = \int_x^{x+1} (x+1-t)f''(t) \, dt $$ is in ${\cal L}^1$ and that can be shown with the Fubini-Tonelli theorem: $$ ...


7

Here is a try: Fix $s \leq t$, then \begin{align*} M_t &= \max \left\{ \sup_{r \leq s} B_r, \sup_{s < r \leq t} B_r \right\} \\ &= \max \left\{M_s, \sup_{r \leq t-s} (B_{r+s}-B_s)+B_s \right\}. \end{align*} The restarted process $W_r := B_{s+r}-B_s$, $r \geq 0$, is again a Brownian motion. If we denote by $M_t^W := \sup_{r \leq t} W_r$ its running ...


7

It suffices to show, that the conditions imply that $P(\{\omega\}) = 0$ for all $\omega \in \Omega$, and thus that $P(\Omega)=0$, contradicting the fact that $P$ should be a probability measure. So from now on consider a fixed $\omega \in \Omega$. Define sets $B_1,B_2,\dots$ in the following way $$B_i := \begin{cases} A_i &, \text{ if $\omega \in A_i$} \\...


7

Claim: For any set $X$ obtainable the boundary is nowhere dense. Proof: Boundary of open set is nowhere dense. Now for the operations: 1) complement - since boundary is the same for a set and its complement, this is automatic 2) union - boundary of a union is subset of union of boundaries, and so it is a subset of a union of two nowhere dense sets, so is a ...


7

You're absolutely right -- the lemma is true without assuming compactness, with an easy proof based on Fubini/Torelli. And you're also right that the reason I assumed compactness was so I could give an elementary proof that doesn't use any measure theory. My intention for my series of manifolds books (Introduction to Topological/Smooth/Riemannian Manifolds) ...


7

Here's an alternative answer not using dominated convergence. Recall that $S \mapsto \int_S |f|$ is a measure so by disjoint additivity, $$\sum_{n = 1}^\infty \int_{[n, n + 1)} |f(x)| = \int_{[1, \infty)} |f(x)| \leq \int_{\mathbb{R} } |f(x)| < \infty $$ which implies that $\lim_{n \to \infty} \int_{[n, n +1)} |f(x)| = 0$ since otherwise, the sum would ...


7

This is most easily understood when $N = 1$. Suppose $A \subset \mathbb{R}^m$ is a bounded domain and $f: A \rightarrow \mathbb{R}$ is a smooth function whose gradient is everywhere nonzero. Its level sets are non-intersecting hypersurfaces. Suppose first that you want to compute the volume of the $A$ in terms of the surface areas of the hypersurfaces. ...


7

Since $fg\geq 1$, you have $\sqrt{f(x)\cdot g(x)} \geq 1$ for all $x$, and so $$ 1 \leq \int_X \sqrt{f}\cdot \sqrt{g} d\mu \leq \sqrt{\int_X f d\mu }\cdot \sqrt{\int_X g d\mu } $$ where the second inequality is Cauchy–Schwarz; squaring both sides gives the inequality.


6

This result is only true for $p \in [1, \infty)$. Let me start by giving a counterexample for the case $p = \infty$. Let $E = [0,1]$ with its Borel $\sigma$-algebra and the Lebesgue measure. Let $A_n f := 1_{[0, 1-\frac1n]} f$ and $B = \operatorname{Id}$. It is easy to check that these are both contractions in $L^q$ for every $q \in [1,\infty]$ and to check ...


6

My answer is very similar to the solution described in Folland's real analysis book. Consider the map $L : C([a, b]) \to \mathbb{R}$ given by $$L(f) = f(a).$$ Note that, here $C([a,b])$ is the space of continuous functions on $[a,b]$. We note that $C([a,b])$ can be viewed as a vector subspace of $L^\infty([a,b])$. Remark: Technically, $C([a,b])$ is not a ...


6

We have \begin{align} I_n &= \int_0^1 \sqrt{\frac{1}{x} + n^2 x^{2n}} \mathrm{d} x \\ &\le \int_0^1 \left(\sqrt{\frac{1}{x}} + nx^n\right) \mathrm{d} x \\ &\le \int_0^1 \left(\sqrt{\frac{1}{x}} + (n+1)x^n\right) \mathrm{d} x \\ &= \left(2\sqrt{x} + x^{n+1}\right)\Big\vert_0^1 \\ &= 3. \end{align} Also, we have \begin{align} I_n &= \...


6

As explained in the last paragraph here, if you combine ZF set theory with the assumption that the axiom of choice is false, the Banach-Tarski paradox becomes undecidable rather than refutable. Indeed, ZF plus something weaker than AC called the ultrafilter lemma renders the BT a theorem; it doesn't need full ZFC. For more details, see the 1991 paper that ...


6

Open intervals $(a,b)$ with $a$ and $b$ rational form a countable class generating the Borel sigma algebra of $\mathbb R$.


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