# Tag Info

1

The Riemann integral of a continuous function on close interval coincides with its Lebesgue integral. $\int_a^{1-a} |f(x)| dx$ is a Riemann integral for each $a \in (0,\frac 1 2)$. And $\int_a^{1-a} |f(x)| dx \to \infty$ as $a$ decreases to $0$ since $|ln x| \leq |\ln a|$ on $(a,1-a)$ and $\int_a^{1-a} \frac 1 {x^{2}}dx\to \infty$ by direct calculation ...

1

Typically, to show $f$ is measurable you need to show that $$\{x: f(x)< c\}$$ is measurable for all $c$. (Equivalently, one can change $<$ to $\leq$, $\geq$, or $>$). It should be pretty easy to break down $c$ into cases and show that these sets amount to intervals, which are measurable.

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You tagged your question with [functional-analysis] so a functional analytic answer might be appropriate. Given a Banach space $X$, and any linear subspace $S$ of the topological dual $X'$ of $X$, one has that $S$ is dense in the weak$^*$ topology (i.e. the topology of pointwise convergence) iff the only vector $x$ in $X$ such that $x'(x) = 0,$ for every $... 1 Theorem: If$f:X\to Y$is a continuous mapping between compact metric spaces and$f(X)=Y$, then there is a Borel set$\mathscr{B}\subset X$such that$f|_{\mathscr{B}}$is one-to-one,$f(\mathscr{B})=Y$and$f^{-1}:Y\to\mathscr{B}$is Borel. Proof: Suppose, initially, that$X \subset [0,1]$. Let$g:Y \to X$be defined by $$g(y) = \inf\{x \in X: f(x)= y \} \,... 3 This is a special case of some high-powered machinery in Descriptive Set Theory. I tried for a while to find an easier approach, but I was unable to. Theorem: (cf. Theorem 6.3 here, or Section 5.2 of Srivastava's "A Course on Borel Sets") Let X and Y be polish spaces, and F : X \to 2^Y a function so that F(x) is \sigma-compact for each ... 0 Let g_n=\inf \{f_k: k \geq n\}. Verify that g_n is non-negative, measurable and increases to f. By Monotone Convergence Theorem \int g_n \to \int f. Also g_n \leq f_n so \int f =\lim \int g_n \leq \lim \int f_n. 0 For a nonnegative measurable function, the integral \int_X f\, d\mu is defined as the supremum of integrals \int_X \varphi \, d\mu where \varphi ranges over all simple functions with 0 \leqslant \varphi \leqslant f. Since \underset{X}\inf f \cdot \chi_X is a simple function satisfying this condition, the first inequality is immediate, viz.$$\... 0 It is not to be understood has "a closed subset in the subset topology" but has "find a closed subset that contains no rational number such that its Lebesgue measure is greater that$\frac{3}{4}$". Otherwise, the answer is trivia : take$[0,1]\setminus \mathbb{Q}$. Let$\{q_n\}_{n\geqslant 1}$be the sequence of all rational numbers in$[...

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Your argument accomplishes more than just making a set $F$ closed in the subspace topology of $E$. It creates a subset $F\subset E$ which is closed in $\mathbb R$ (and so certainly also in the subspace topology). Namely, a countable union $U\subset [0,1]$ of smaller-and-smaller open intervals around rationals is open in $\mathbb R$, and can be made to have ...

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I will try to answer your question for indicator functions $\chi_S$. This may give some indication how the question could be answered for simple functions, and perhaps arbitrary functions. For any given partition $E = \bigsqcup_{k=1}^n E_k$, notice $\sup_{x\in E_k} \chi_S(x)$ is $1$ if $E_k$ hits $S$ and $0$ otherwise. Let $A$ be the union of all the $E_k$'s ...

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I don't know exactly what you mean by "finding" $\sigma(J)$, since in both cases you mentioned, the $\sigma$-algebra will have infinite cardinality. In fact this is exactly the purpose of the notation $\sigma(J)$, which is to describe a large $\sigma$-algebra without listing its elements. To get a feel for the measurable sets, start by thinking of ...

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I will use the following well known result: If $g$ is integrable on $\mathbb R^{d}$ then, for any $\epsilon >0$, we can find a continuous function $h$ such that $\int |g-h| <\epsilon$. For each $n$ there exists a continuous function $\phi_n$ such that $\phi_n(x)=1$ if $\|x||<n$ and $0$ if $\|x|| >n+1$. Consider $\phi_n \arctan f$. This function ...

1

Consider this set $M = \{A\in\cal{B}(\mathbb{R})| A\times \mathbb{R} \in \cal{B}(\mathbb{R}^2)\}.$ If $A$ is open, $A\in M$. What else can you say about $M$?

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There is no such $\mu$. If $\mu$ exist then $p_n \to p$ uniformly on $[0,1]$ would imply that $p_n'(0) \to p'(0)$ (since $\int_0^{1}[p_n(x)-p(x)]d\mu (x) \to 0$). Take $p_n(x)=\frac {(x-1)^{n}} n$ for a counter-example.

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Lebesgue measure is the unique (Borel) measure which is translation invariant, finite on compact sets and attains 1 on the unit cube (without this last one, you can scale the Lebesgue measure by some $c>0$ and still have a translation invariant measure). Here is a sketch of the proof. As far as I know, you need finiteness on compact sets, so that the ...

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$$\chi_E S = \sum\limits_{i=1}^{n} \alpha_i \chi_E \chi_{A_i}$$ $$=\sum\limits_{i=1}^{n} \alpha_i \chi_{E \cap A_i}.$$ This is a simple function. Simply omit the terms where $E \cap A_i$ is empty. This implies that $\int \chi_E d\mu=\sum\limits_{i=1}^{n} \alpha_i \mu (\chi_{A_i}\cap E)$. It makes no difference whether $0$ is in the range or not.

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Hint: Let $f_n(x)=-x^{-\frac{n-1}{n}}$.

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How about something like this: If $x$ is rational and $x=a/b$ in lowest terms, let $f(x) = 1/b$. If $x$ is irrational, let $f(x) = 34$. This is supposed to have the properties: $f(x) > 0$ for all $x$, and $\inf\{f(x), a < x < b\} = 0$ for all nonempty open intervals $(a,b)$.

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Denote by $f:[0,1] → \mathbb R$ the function in question: $$f (x) := \mathbb P \big((0,x]\big).$$ Firstly, I am not convinced your formula at the end of the question holds: i.e., $$f(x) = \sum_{n=1}^∞ \frac{x_n}{2^n},$$ since the right hand side equals $x$! And this is far too rigid: indeed, $f(1/2) = p$ is not necessarily equal to $1/2$. Part 1 Let's take ...

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\begin{align} & \{x\in X: f(x)>0\} \\[8pt] = {} & \bigcup_{n\,\in\,\mathbb N} \left\{x\in X: \tfrac 1 {n+1} \le f(x)< \tfrac 1 n \right\} \cup \{x\in X: f(x)\ge1\}. \end{align} The measure of this union is positive only if the measure of one of the terms is positive. So you need to show that implies the integral is positive.

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Simple functions have a finite amount of values, so if $f$ is a simple function then $\text{im}(f)=\{a_1,\ldots,a_n\}$. If $g$ is another simple function then $\text{im}(g)=\{b_1,\ldots,b_m\}$. We know $X=\cup_{l=1}^m g^{-1}(\{b_l\})$, since every $x\in X$ gives us one of those values as its image under $g$; and this union is of pairwise disjoint sets, since ...

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This is a proof using 'linear trasformations', not in the sense you cited : Is sufficient prove it for the hyperplane $P_{0} := \{(y_{1},\cdots,y_{d}) : y_1 = 0\} \cong \mathbb{R}^{d-1} \subset \mathbb{R}^{d}$, using the invariancy of isometry of Lebesgue measure, the completeness of measure. Given $\epsilon > 0$ let $Q_{n} = [-\frac{\epsilon}{(2n)^{d-1}2^... 1 I use my notation.$E$is your$A$and$E_j$'s are your$B_j$'s.$\ge$: Observe that for$E\in\mathcal{A}$, $$|\mu|(E)\leq|\mu^+(E)|+|\mu^-(E)|=|\mu|(E)$$ By countable additivity, $$|\mu|(E)=\sup\left\{\sum_{j=1}^{n}|\mu|(E_j):\ldots\right\}\ge\sup\left\{\sum_{j=1}^{n}|\mu(E_j)|:\ldots\right\}$$$\leq$: Apply Hahn decomposition,$|\mu|(E)=|\mu|(E\cap A)+|\mu|...

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I think the important thing to note is that $F_1^{-1}(x)$ and $F_2^{-1}(x)$ both just look like $F(x)$ flipped around the line $y=x$; in particular, they're exactly the same except where $F$ plateaus, and at these points $F_1^{-1}$ and $F_2^{-1}$ have to make a vertical jump, $F_1^{-1}$ is just the flipped version of $F$ where we pick the bottom of the ...

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$F_1^{-1},F_2^{-2}$ are somekind of "generalised" inverses of $F$. Firstly, we'll show that $F_1^{-1}(\omega) \le t \iff \omega \le F(t)$ => Suppose that $F_1^{-1}(\omega) \le t$. By definition of infimum and non-decreasing property of $F$, it means that for $h>0$ we have $F(t+h) \ge \omega$, but then by rightcontinuity of $F$ (and again, non-...

1

Since Lebesgue's integral extend Riemann's, every convergence theorem for Lebesgue integral also holds for Riemann integral as long as the sequence of functions involved, and its limit, are Riemann integrable functions. However, the pointwise limit of Riemann integrable functions might not be Riemann integrable, so stating such results in the context of ...

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Let $\mu$ be the Lebesgue measure on $[0,1]$. Let $I’=\{x\in [0,1]: x\mbox{ has a unique decimal expansion}\}$. Since $[0,1]\setminus I’$ is a set of numbers admitting a finite decimal expansion, it is countable, so $\mu(I’)=\mu([0,1])=1$. For each digit $i$, let $I’_i=\{x\in I’: \mbox{ a decimal expansion of$x$has a digit$i$at its$k$-th place}\}$. It ...

1

Since $f \chi_{E_n} \geq 2n\chi_{E_n}$ we get $\int_{E_n} f \geq \int_{E_n} 2n = 2n \mu (E_n)$ and $\int_{E_n} f \to 0$ so $\lim n \mu(E_n)=0$.

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Every measurable function (in particular, $f$) is a pointwise limit of a sequence $(s_n)$ of measurable functions which are finite-valued (this is certainly a theorem in Stein and Stakarchi's text, though I don't know where). Now $(s_n\chi_{E_\alpha})$ converges pointwise to $f\chi_{E_\alpha}$, and any pointwise limit of a sequence of measurable functions ...

1

As you suggested, define for $t \in [0,1]$ $$A:=A(t) := \bigcup_{n \in \mathbb{N}} (q_n-t2^{-n},q_n+t 2^{-n}).$$ If we set $$f(t) := \lambda(A(t)), \qquad t \in [0,1],$$ then $f$ is continuous and satisfies $f(0)=0$, $f(1)\geq1$. By the intermediate value theorem, there exists for any $m \in [0,1]$ some $t \in [0,1]$ with $\lambda(A(t))=f(t)=m$; the set $A(... 1 You can take the complement (in$[0,1]$) of a fat Cantor set with measure$1-m$. If$K$is such a fat Cantor set, then: since$K$has Lebesgue measure$1-m$, the Lebesgue measure of$[0,1]\setminus K$is$m$; since$\mathring K=\emptyset$,$[0,1]\setminus K$is dense; since$K$is compact,$K$is a closed subset of$[0,1]$, and therefore$[0,1]\setminus K$... 1 Take an enumeration$\mathcal E:=\{q_n\}_{n\in\mathbb N}$of$\mathbb Q\cap(0,1)$satisfying that for each for each$n$we have that $$0\leq q_n-\frac{m}{M_\mathcal E}m2^{-n-1}<q_n+\frac{m}{M_\mathcal E}m2^{-n-1}\leq1,$$ where $${M_\mathcal E}=\mu\left(\bigcup_{n\in\mathbb N}\left(q_n-m2^{-n-1},q_n+m2^{-n-1}\right)\cap[0,1]\right).$$ Note that if$q_1=\...

1

Consider a set $$A=\bigcup_{n=1}^{\infty} \left(\sqrt{n} ,\sqrt{n+\frac{1}{n}}\right).$$ Since $$-\sqrt{n} +\sqrt{n+\frac{1}{n}}=\frac{\frac{1}{n}}{\sqrt{n} +\sqrt{n+\frac{1}{n}}} \leq \frac{1}{2(\sqrt{n})^3}$$ the set $A$ has finite measure an is obviously Lebesgue measurable. Now consider a function $h:\mathbb{R}\to\mathbb{R}$ defined belov: $$h(x) =\begin{... 2 The last one is not true as well. Take f s.t. f(x)=e^x when x>0 (you can prolonge it as an homeomorphism over \mathbb R). Take$$A=\bigcup_{n=2}^\infty \underbrace{\left(n,n+\frac{1}{n^2}\right)}_{=:A_n}.$$Then m^*(A)<\infty  but m^*f(A))=\infty  because m^*f(A_n)=e^n\left(e^{\frac{1}{n^2}}-1\right), and$$\sum_{n\geq 2}e^n\left(e^{\frac{...

3

We construct $\{I_k\}_{k=1}^{\infty}$ inductively. Enumerate the natural numbers with $m$, beginning with $m=1$. Let $\epsilon=\frac{1}{m}$. Find inductively, using the Vitali Covering Lemma, a finite disjoint collection of closed intervals, $\{I_{mk}\}_{k=1}^{n_m}$, for which $$m^* (E_m) < \frac{1}{m},$$ where we define $E_1= E \sim \bigcup_{k =1}^{n_1} ... 1 There exist sets$E,F$of measure$0$such that$f(x) \leq 1$if$x \notin E$and$f(x) \geq 1$if$x \notin F$. Let$A=E \cup F$. Then$A$has measure$0$and$f(x) = 1$if$x \notin A$(because$x \notin A$implies$x \notin E$and$x \notin F$). Now let$D=\{x: f(x)=1\}$. Then$D$is a Lebesgue measurable set and$f=1_D$almost everywhere. To complete ... 2 Suppose otherwise. Then for each$N$you can find Lebesgue measurable sets$A_N$and$B_N$such that$A_N \subseteq E \subseteq B_N$and$m(B_N \setminus A_N) < 1/N$. Without loss of generality we may assume that$A_1 \subseteq A_2 \subseteq A_3 \subseteq \dots$and$B_1\supseteq B_2 \supseteq B_3 \supseteq \dots$(indeed, if necessary we can replace$...

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I would recommend proving the contrapositive: show that if for any $\eta > 0$ there exists Lebesgue measurable sets $A$ and $B$ such that $A \subseteq E \subseteq B$ and $m(B\backslash A) \leq \eta$, then $E$ is measurable. Hint: all sets of Lebesgue outer measure $0$ are measurable.

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Not true. Let $x_k=x$ for all $n$ and $A_k=(t-\frac 1 k, t +\frac 1 k)$. Then $\frac 1 {\mu (A_k)} \int_{A_k} |g(x_k(s))-f(x_k(s))|ds \to |g(x(t))-f(x(t))|$ as $k \to \infty$.

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Note: This is not actually an answer, only a place to share some preliminary observations in the hopes that it will be useful for others. Idea for simplifying notation/proofs: Instead of using closed intervals in the question, one could use only half-open intervals (for example, of the form $[a,b)\subset [0,1)$). The problem statements should be equivalent ...

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We prove that the equation holds for each open interval and then use the conclusion of a). Let $L$ be an open interval and $L=(L\cap I)\cup (L\cap I^{c})$. Since $I$ is measurable, we have $$\lambda ^{*}(L)=\lambda ^{*}(L\cap I)+\lambda ^{*}(L\cap I^{c})$$ Also, we have \lambda ^{*}(I\cap L)\leq \lambda ^{*}(B\cap I\cap L)+\lambda ^{*}(B^{c}\cap I\cap L)(1)...

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Without the hypothesis $\int g<\infty$ there are counter examples: Let's take $g(x)=\dfrac 1x\$ and $\ g_n(x)=g(x)\ \chi_{[\frac 1n,1]}$ Now with $f_n(x)=n(1-x)^n\ \chi_{[\frac 1n,1]}\$ and $\ f(x)=0$

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This is false if there are no integrability assumptions. Let $f_n=n\chi_{(0,\frac 1 n)}, g_n\equiv n, f(x)=0$ and $g(x)=\infty$ for all $x$. Then the hypothesis is satisfied. But $1=\int f_n$ for all $n$ and $\int f=0$. If $g$ is integrable then an application of Fatou's Lemma to $(g_n-f_n)$ gives $\lim \sup \int f_n \leq \int f$. Another application of ...

2

Here is a sketch for a solution which uses the assumption that $\int g <\infty$ (leave a comment if you want more details for the solution): You have already shown $\int f \leq \liminf \int f_n$. If we apply Fatou to the nonnegative sequence $g_n-f_n$, then one can obtain $\int f \geq \limsup \int f_n$ after using the other assumptions. But what if $\int ... 1 Your answers for 1), 2),4)and 5) are correct. For 3) just note that you can approximate$m^{*}(A)$by a closed set$C$contained in$A$and consider the compact set$K=C\cap[ -N,N]$with large enough$N$. I hope you can write out a proof using this hint. 0 Unless I'm misinterpreting something, this is false. Let$\Sigma_1$be the Borel measurable sets and$\Sigma_2$be the Lebesgue measurable sets. Let$\mu_1^*$and$\mu_2^*$both be the Lebesgue outer measure and let$\mu_1$be Lebesgue measure (restricted to Borel sets only) and let$\mu_2$also be Lebesgue measure. Then all the properties listed are true,... 1 Hint/comments to get started: A function$f$is integrable on a set$E$if$\int_E|f|$is finite. For$x$near$0$,$f(x)$looks like the constant function$1$, so you can focus on$f(x)$for values of$x$larger than$1$, say. So, is$\int_{1}^\infty|f|$finite? Well, if we ignore the$\cos(x)$term, then$|f(x)| = 1/\sqrt{x+1}$, which is basically$1/\sqrt{...

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This does not answer the explicit questions but the first implicit one above. Other important examples of nonmeasurable sets are Bernstein's. Those are based on the inner regularity of measure: Every non null set includes a non null (and, a fortiori uncountable) closed set. A Bernstein set $B\subset \mathbb{R}$ satisfies that $F\nsubseteq B, B^c$ for every ...

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I consider the case that $n=1$ only. Let $\mathcal{L}$ be the $\sigma$-algebra of all Lebesgue measurable subsets of $\mathbb{R}$. $\mathcal{L}\subseteq\mathcal{M}$ follows from the outer-regularity of Lebesgue measure (which can be found in any real analysis textbook). To show $\mathcal{M}\subseteq\mathcal{L}$. Let $A\in\mathcal{M}$. For each $n\in\mathbb{N}... 0 Here is an update: Now, I'm taking a class on real analysis, and it has been pointed out to me (thanks Alan!): in order to cover$B$with a fininte number of open boxes with an error of at most$\varepsilon\$, we don't need to merely rely on the definition of the outer measure to tell us that such a collection exists. Rather, we can actually write down the ...

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