Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $(X,\mathcal{M},\mu)$ be a measure space, $Y$ a separable Banach space, and $L_{Y}$ the space of all $(\mathcal{M},\mathcal{B}_{Y})$-measurable maps from $X$ to $Y$ (where $\mathcal{B}$ denotes the Borel $\sigma$-algebra). Let $F_{Y}$ be the set of maps $f:X\rightarrow Y$ of the form $f(x)=\sum_{j=1}^{n}\chi_{E_{j}}(x)y_{j}$ where $n\in\mathbb{N},y_{j}\in Y,E_{j}\in\mathcal{M}$, and $\mu(E_{j})<\infty$. If $f\in L_{Y}$, since $y\mapsto||y||$ is continuous, $x\mapsto||f(x)||$ is $(\mathcal{M},\mathcal{B}_{\mathbb{R}})$-measurable, and we define $||f||_{1}=\int||f(x)||\;d\mu(x)$. FInally, let $L_{Y}^{1}=\{f\in L_{Y}:||f||_{1}<\infty\}$.

This is a 6-part question in Folland's Real Analysis (problem 5.16). I've done the first two parts (which I will state below for reference) and I'm stuck with the third.

(a) $L_{Y}$ is a vector space, $F_{Y}$ and $L_{Y}^{1}$ are subspaces of it, $F_{Y}\subset L_{Y}^{1}$, and $||\cdot||_{1}$ is a seminorm on $L_{Y}^{1}$ that becomes a norm if we identify two functions that are equal a.e.

(b) Let $\{y_{n}\}_{n=1}^{\infty}$ be a countable dense set in $Y$. Given $\varepsilon>0$, let $B_{n}^{\varepsilon}=\{y\in Y:||y-y_{n}||<\varepsilon||y_{n}||\}$. Then $\bigcup_{n}B_{n}^{\varepsilon}\supset Y\setminus\{0\}$.

(c) If $f\in L_{Y}^{1}$, there is a sequence $\{h_{n}\}\subset F_{Y}$ with $h_{n}\rightarrow f$ a.e. and $||h_{n}-f||_{1}\rightarrow 0$. (With notation as in (b), let $A_{nj}=B_{n}^{1/j}\setminus\bigcup_{m=1}^{n-1}B_{m}^{1/j}$ and $E_{nj}=f^{-1}(A_{nj})$, and consider $g_{j}=\sum_{n=1}^{\infty}y_{n}\chi_{E_{nj}}$.)

share|cite|improve this question

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.