It is rather straight forward to show that $L_p$ is complete for $p\geqslant 1$, but I am having trouble showing the same thing when $p<1$. For the former case I have shown that every absolutely convergent sequence converges by constructing a a function in $L_p$ but bigger than the series and used the dominated convergence […]

Could someone explain the intuition behund the Hausdorff Measure and Hausdorff Dimension? The Hausdorff Measure is defined as the following: Let $(X,d)$ be a metric space. $\forall S \subset X$, let $\operatorname{diam} U$ denote the diameter, that is $$\operatorname{diam} U = \sup \{ \rho(x,y) : x,y \in U \} \,\,\,\,\, \operatorname{diam} \emptyset = 0 $$ […]

Let $B$ be a bounded Borel set of $\mathbb{R}$, Show that if $A$ is a finite union of disjoint intervals, the Lebesgue measure of $A\triangle B$ can be arbitrarily small. Also show that this remains true as long as $B$ has finite Lebesgue measure.

Let $X$ be the space of measurable functions $f:[0,1] \rightarrow \mathbb{R}$. I want to find out whether this space is complete under the metric $d(f,g):= \int_{[0,1]} \frac{|f-g|}{1 + |f-g|}$. Does anybody here have an idea how to tackle this problem?

Let’s start off with two measure spaces, $ (X,\mathcal{A},\mu) $ and $ (Y,\mathcal{B},\nu) $, and suppose that we want to form their product measure space. It can happen that more than one product measure exists: One has the measures arising from the Carathéodory construction on the $ \sigma $-algebra $ \mathcal{A} \times \mathcal{B} $ corresponding […]

Let $\mathbb K$ be $\mathbb R$ or $\mathbb C$. Let $(X, \mathcal M, \mu)$ be a measure space and let $F$ be a Banach space over $\mathbb K$. I would like to define an integral of a function $f:X \rightarrow F$ that satisfies suitable conditions. My method is motivated by Bourbaki’s Integration. They assume that […]

Set of rational numbers $\mathbb{Q}$ is measure $0$. I approach this question by two sides. (First) Like here Showing that rationals have Lebesgue measure zero., $$ \mu(\mathbb Q) = \mu\left(\bigcup_{n=1}^\infty \{q_n\}\right) = \sum_{n=1}^\infty \mu(\{q_n\}) = \sum_{n=1}^\infty 0 = 0. $$ (Second) Using the definition of Lebuesgue measure. let’s order $\mathbb{Q}=\bigcup_{i=1}^{\infty}\left\{ r_{i}\right\} $. For given $\epsilon>0$, […]

I’m new to formal writing, so please be patient :). Let $S=\left\{ s \subset [0,1] : |s|=n \right\}$, and let $x$ be a probability distribution over the elements of $S$. Namely, $x$ is a distribution over all sets of size $n$ (for a constant $n$) of elements from $[0,1]$. Define $\mu: 2^{[0,1]}\rightarrow [0,n]$ such that […]

There are several variants of dominated convergence theorem. The standard one requires $f_n \to f$ a.e. and $|f_n|\leq g$ a.e. where $g$ is integrable. It can be weakened to only convergent in measure, if we impose $\sigma$-finiteness to the measure, c.f. Generalisation of Dominated Convergence Theorem. However, we know if $g$ is integrable, $N(g) = […]

Let $(X,\mathbb X,\mu)$ be a measure space and let $(E_n)$ be a sequence in $\mathbb X$. Show that $$\mu(\lim\inf E_n)\leq\lim\inf\mu(E_n).$$ I am quite sure I need to use the following lemma. Lemma. Let $\mu$ be a measure defined on a $\sigma$-algebra $\mathbb X$. If $(E_n)$ is an increasing sequence in $\mathbb X$, then $$\mu\left(\bigcup_{n=1}^\infty E_n\right)=\lim\mu(E_n).$$ […]

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