Possible Duplicate: Convergence a.e. and of norms implies that in Lebesgue space I am trying to show that if $$ \int_X |f_n|d\mu \to \int_X|f|d\mu $$ where $f$ and all the $f_n$ have finite integral and $f_n \to f$ pointwise, then $$ \int_X |f_n-f|d\mu \to 0. $$ I worked out a proof in the case that […]

I am having some conceptual difficulties with almost everywhere (a.e.) convergence versus convergence in measure. Let $f_{n} : X \to Y$. In my mind, a sequence of measurable functions $\{ f_{n} \}$ converges a.e. to the function $f$ if $f_{n} \to f$ everywhere on $X$ except for maybe on some set of measure zero. However, […]

Given a sequence $(\mu_n)_{n\in \mathbb N}$ of finite measures on the measurable space $(\Omega, \mathcal A)$ such that for every $A \in \mathcal A$ the limit $$\mu(A) = \lim_{n\to \infty} \mu_n(A)$$ exists. I want to show that $\mu$ is a measure on $\mathcal A$. What I managed to figure out: $\mu$ is monotone, additive and […]

We know that if $ f: E \to \mathbb{R} $ is a Lebesgue-measurable function and $ g: \mathbb{R} \to \mathbb{R} $ is a continuous function, then $ g \circ f $ is Lebesgue-measurable. Can one replace the continuous function $ g $ by a Lebesgue-measurable function without affecting the validity of the previous result?

After several hours of struggling, I’ve been unable to solve the following problem Let $X,Y: (\Omega, \mathcal{S}) \rightarrow (\mathbb{R}, \mathcal{R})$ where $\mathcal{R}$ are the Borel Sets for the Reals. Show that $Y$ is measurable with respect to $\sigma(X) = \{ X^{-1}(B) : B \in \mathcal{R} \}$ if and only if there exists a function $f: […]

I’m working through a proof and one of the comments is that for a function $f\in L_p (\mathbb{T})$: $$\lim_{t\to 0}\;\|f(\cdot + t) – f\|_p = 0.$$ How do I prove it? I think it is intuitively clear if $f$ is a step function, but what about for an arbitrary $p$ integrable function?

A function $f:[0,1]\rightarrow\mathbb{R}$ is called singular continuous, if it is nonconstant, nondecreasing, continuous and $f^\prime(t)=0$ whereever the derivative exists. Let $f$ be a singular continuous function and $T$ the set where $f$ is not differentiable. Question: Is $T$ nowhere dense? Examples: A classical example of such a function is the so-called devil’s staircase, obtained as […]

In the following, consider the Lebegue measure in $\mathbb{R}^d$. Consider $E\subseteq \mathbb{R}^d$ measurable, with $0\lt m(E)\lt\infty$, such that any measurable subset $F$ of $E$ satisfies $m(F)=m(E)$ or $m(F)=0$. What can we say about $E$? Does there exist such a set?

From Baire category theorem, we see that $\mathbb{Q}$ can not be a $G_{\delta}$. But consider the following construction: Let us consider $\mathbb{Q}\cap [0,1]$, putting all the elements in the set in a sequence, denoted $\{a_n\}$. We define $$V_i=\bigcup_{j}[a_j-1/2^{i+j},a_j+1/2^{i+j}]\cap [0,1].$$ Notice that $\mathbb{Q}\subset V_i$. So we define $$V=\bigcap_{i} V_i.$$ We have $\mathbb{Q}\subset V$, and $V$ is […]

I am reading a theorem about measurability of vector-valued functions in a note on functional analysis: Theorem 3.6.1. If $X$ is a separable, metrizable locally convex space, $(\Omega, \Sigma, \mu)$ is a $\sigma$-finite measure space, and $f : \Omega \to X$ is weakly measurable, then $f$ is strongly measurable. The proof begins We can restrict […]

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