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 $...


2

Assume $f\in L^{\infty}$. Then, $|f|\leq \|f\|_{\infty}$ almost everywhere and we get that $$ \|f\|^p_p=\int |f|^p\textrm{d}\lambda\leq \int \|f\|_{\infty}^p\textrm{d}\lambda=\lambda(\Omega)\|f\|^{p}_{\infty}, $$ implying that $\limsup_{p\to\infty}\|f\|_p\leq \|f\|_{\infty}$. Similarly, for any $\|f\|_{\infty}>\varepsilon>0$, we have that $|f|\geq 1_{\...


1

Since the Integral is linear it follows $$ \begin{align} \int_X f d \mu = -\int_X (-f) d \mu &= -\sup \{\mathcal{L}(-f, P) \mid P \text { is an } \mathcal{S} \text { -partition}\}\\ &= - \sup \{-\ \mathcal{U}(f, P) \mid P \text { is an } \mathcal{S} \text { -partition}\}\\ &=\inf \{\ \mathcal{U}(f, P) \mid P \text { is an } \mathcal{S} \text { -...


1

First, by Jensen's inequality: $$(Mf(x))^p=\sup_{r>0}(\frac{1}{m(B(x,r))}\int_{B(x,r)}|f(y)|\,dy)^p\leq \\\sup_{r>0}\frac{1}{m(B(x,r))}\int_{B(x,r)}|f(y)|^p\,dy=M|f|^p(x)$$ Thus: $$m(\{x:Mf(x)>\lambda\})=m(\{x:(Mf(x))^p>\lambda^p\})\leq m(\{x:M|f|^p(x)>\lambda^p\})\leq \\\frac{3^n}{\lambda^p}|||f|^p||_1=\frac{3^n}{\lambda^n}||f||_p^p$$


1

The $\sigma$-algebra is, as you wrote, all subsets of $\mathbb N$, so that says all such subsets are measurable, and every function on $\mathbb N$ is measurable. Integration with respect to this measure is summation with the weights $w_k$: $\int_{\mathbb N} f\; d\nu = \sum_{k \in \mathbb N} w_k f(k)$. Integrable functions are those such that $\sum_{k \in \...


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.


1

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 \} \,...


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