Articles of measure theory

Does simply-connected imply measureable?

The famous examples of non-measureable sets involve a sophisticated selections of points from a ball (or another object). This raises the following question: if a certain object in a Euclidean space is simply-connected, does it imply that it is Lebesgue-measureable? (BACKGROUND: I am doing a geometry-related research, where most shapes are simple polygons and circles. […]

find the limit: $\lim_{n\to\infty}\int_{0}^{\infty} \frac{\sqrt x}{1+ x^{2n}} dx$

Calculate the following limit $$\lim_{n\to\infty}\int_{0}^{\infty} \frac{\sqrt x}{1+ x^{2n}} dx$$ I tried to apply dominated convergence theorem but I could not find the dominating function. even I broke down the integration from $0$ to $1$ and $1$ to infinity. then found only integration from $0$ to $1$ is possible. Do you have any ideas?

Calculating a simple integral using abstract measure theory

Let $(X,\mu)$ be a measure space. For a measurable function $\iota:X\to X$, we have for all measurable $E\subset X$, $$\int_{\iota^{-1}(E)}g\circ\iota \, d\mu = \int_{E}g \, d(\iota^{*}\mu),$$ where by definition $\iota^{*}(\mu)[A] = \mu(\iota(A))$. I’m trying to reconcile this with the more elementary process I used to refer to as “$u$ substitution”. Consider the case of $X […]

Show a locally integrable function vanishes almost everywhere

Let $u\in L^{1}_{loc}(\Omega):=\{f:\Omega \to \mathbb R\;| \int_{K}|f(x)|dx<\infty,\;K\subset\Omega\; \mathtt{compact}\} $, where $\Omega\subset\mathbb{R}^{n}$, and let $\phi$ be a test function on $\Omega$. Show that $$ \left(\forall\phi\in \mathbb{D}(\Omega):\int\limits_{\Omega}u(x)\phi(x)dx=0\right)\implies \left(u =0\; \mathtt{a.e.}\right) $$ Here’s how I started. Take some $\phi$ in $\mathbb{D}(\Omega)$. Then there is a compact set $K$ such that $\{x: \phi(x)\neq 0\}\subset K$. Since $\phi$ is continous, […]

Are there Lebesgue-measurable functions non-continuous almost everywhere?

My intuition keeps telling me that being continuous Lebesgue-almost everywhere is highly restrictive and that being measurable is not. But I’ve not been able to come up with a not continuous a.e. function e.g. $[0,1] \longrightarrow \mathbb{R}$. So Are there not continuous a.e. functions? Are there Lebesgue-measurable ones?

Preservation of martingale property when changing to a product space.

Let the process $M=(M_t, t\ge 0)$ be a martingale on the probability space $(\Omega_1, \mathcal F_1, P_1)$ with respect to the natural filtration of $M$. Let $X=(X_t, t\ge 0)$ be a process on the probability space $(\Omega_2, \mathcal F_2, P_2)$. Let $W=(M,X)$ be the coupled stochastic process on the probability space $(\Omega_1\times\Omega_2, \mathcal F_1\times\mathcal F_2, […]

Radon-Nikodym Decomposition

Consider $\mathcal{M}([0,1])\equiv C([0,1])^*$ (topological dual) then the Radon-Nikodym decomposition gives us $$\mathcal{M}([0,1])=AC([0,1])\oplus \mathcal{S}[0,1]$$ Where $AC([0,1])$ denotes the space of measures that are absolutely continuous with respect to Lebesgue measure and $\mathcal{S}[0,1]$ are the singular ones. Show that both spaces in this decomposition are not closed in the weak$^\ast$ topology. To show that the first is […]

$L_p$ complete for $p<1$

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 […]

Hausdorff Measure and Hausdorff Dimension

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 $$\operatorname{diam} U = \sup \{ \rho(x,y) : x,y \in U \} \,\,\,\,\, \operatorname{diam} \emptyset = 0 $$ […]

Approximating Borel sets by finite unions of intervals

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.