I’ve noticed two different kinds of definitions for a Measurable Function. In Folland’s Real Analysis Modern Techniques: If $(X, \mathcal {M})$ and $(Y, \mathcal {N})$ are measurable spaces, a mapping $f: X \to Y$ is called $(\mathcal {M}, \mathcal {N})$-measurable, or just measurable when $\mathcal {M}$ and $\mathcal {N}$ are understood, if $f^{-1}(E) \in \mathcal […]

If $C$ is the Cantor Set, I am asked to show that there exists a continuous bijection, say $f$, that maps $C \to [0,1]$. My best guess thus far has been the Cantor Function, however (using this construction) it doesn’t appear to me to be a bijection, specifically not injective. If this is the case, […]

Let $(X,\mathfrak B(X))$ and $(Y,\mathfrak B(Y))$ be measurable spaces and further let $\mu$ be a measure on $\mathfrak B(X)$ and let $K$ be a kernel, i.e. for any $x\in X$ we have $K_x$ is a measure on $\mathfrak B(Y)$ and the map $x\mapsto K_x(B)$ is $\mathfrak B(X)$-measurable for any $B\in \mathfrak B(Y)$. Let us further […]

Let $A$ be a subset of $\mathbb{R}$ such that $\mu^*(A)=0$, where $\mu^*$ is the Lebesgue outer measure. Prove that if $B=\left\{x^2: x\in A \right\}$, then $\mu^*(B)=0$. Recall that the Lebesgue outer measure of a subset $A$ of $\mathbb{R}$ is defined as $$ \mu^*(A)=\inf \left\{\sum_{i \geq 1} (b_i-a_i): (a_i, b_i) \subseteq \mathbb{R}, A\subseteq \cup_{i\geq 1} (a_i, […]

Assume $A$ is a countable dense set in $\mathbb{R}$, and set $B$ has positive (Lebesgue) measure. Prove that $A+B=\{a+b:a\in A, b\in B\}=\mathbb{R}\backslash N$, where $N$ is a set of measure zero. I haven’t come up with a good idea. Thanks in advance!

Let $ (X,\mu) $ be a standard measure space – so that we may assume that $X$ is the unit interval $[0,1]$ with the Borel $\sigma$-algebra. Consider $X \times X$ with the product measure $\mu \times \mu $ defined on the product $\sigma$-algebra. Let $f$ and $g$ be two functions defined on $X \times X$ […]

Let $f\in L^1(\mathbb{R})$ be of compact support and $\psi(x)=C \exp(-(1-x^2)^{-1})$ where $C$ is chosen so that $\int_{\mathbb{R}} \psi =1$. Show that the convolution $f*\psi(x)=\int_{\mathbb{R}} f(x-y)\psi(y) dy$ is infinitely differentiable.

This question already has an answer here: Preimage of generated $\sigma$-algebra 3 answers Question about Borel sets 3 answers

Is it true that if $X$ is a measure space and $\mu, \nu$ are Borel probability measures on $X$ if $$ \int_X \phi \ d \mu = \int_X \phi \ d \nu \qquad \forall \phi \in C_b(X) \text{ (continuous and bounded functions)} $$ then $$ \mu = \nu \text{ ?}$$ If $E$ is a measureable […]

I need this as lemma. Given the Borel space $\mathcal{B}(\mathbb{R})$. Consider a complex measure: $$\mu:\mathcal{B}(\mathbb{R})\to\mathbb{C}$$ Then one has: $$\int_{-\infty}^{+\infty}e^{it\lambda}\mathrm{d}\mu(\lambda)=0\implies\mu=0$$ How can I prove this?

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