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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_{\...
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 \} \,...