1
$\begingroup$

In this proof i used the following fact (1): "If $f\in L^p(\mu), 1\leq p < \infty$, and let $\epsilon > 0$ then there exist a set $E_\epsilon$ measurable with $\mu(E_\epsilon) < \infty$ such that $\displaystyle\int_{E^{c}}|f|^p d\mu < \epsilon$"

Proposition: Suppouse $\{f_k\} \in L^p(\mu), 1 \leq p < \infty$. Let $B_k(E) = \displaystyle\left(\int_E |f_k|^pd\mu\right)^{\frac{1}{p}} = \left(\int |f_{k}\chi_{E}|^pd\mu\right)^{\frac{1}{p}}$, where $X_{E}$ is the characteristic function of $E$. If $\{f_k\}$ is a Cauchy sequence and let $\epsilon > 0$ prove that exist a set $E_\epsilon$ measurable with $\mu(E_\epsilon) < \infty$ such that $B_n(E^{c}_{\epsilon}) < \epsilon$ $ \forall n \in \mathbb{N}$

Proof: Let $\epsilon$ take $n_0 \in \mathbb{N}$ such that for $n, m \geq n_0 \Rightarrow$ $\displaystyle||f_n - f_{m}||_p < \frac{\epsilon}{2}$

By fact (1) for $i = 1, 2, ..., n_0 -1$ there exist measurable sets $E_1, ..., E_{n_0}$ such that $\displaystyle B_1(E_{1}^{c}) = ||f_1\chi_{E^{c}_1}||_p, ..., B_{n_0}(E_{n_0}^c) = ||f_{n_0}\chi_{E_{n_0}^{c}}||_p < \frac{\epsilon}{2}$.

Take $A = E_{1} \cup ...\cup E_{n_0}$, then $A^c = E_{1}^c \cap...\cap E_{n_0}^c$ and $B_1(A^c),...,B_{n_0}(A^c) < \frac{\epsilon}{2}$; for $n \geq n_0$ $\displaystyle B_n(A^c) = ||f_n\chi_{A^c}||_p \leq ||f_n - f_{n_0}\chi_{A^c}|| + ||f_{n_0}\chi_{A^c}||_p < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$. Therefore $\forall n \in \mathbb{N}$ $B_n(A^c) < \epsilon$

$\blacksquare$

Is there some error? If yes, how should i do?

  • 0
    Do you need to prove (1)? (By the way, where you write $X$ you intend to write $\chi$, which is rendered with \chi in LaTeX.) As for your proof, other than some syntactic errors (for example, missing parentheses), it seems fine, again other than not proving (1), which might not be part of the assignment. – 2017-02-04
  • 0
    Thanks. I have a proof of the fact (1), the sketch is as follows: Let $(X, F, \mu)$ the measurable space for each $n \in \mathbb{N}$ consider the set $E_n = \{x \in X; |f(x)|^p > \frac{1}{n}\}$. We have the sets $E_1 \subset...\subset E_n\subset...$ and $X = \bigcup E_n \cup \{|f|^p = 0\}$, define $f_n := f\chi_{E_n}$ then $f_n \uparrow f$ with $f_{n-1} \leq f_{n}$; by the theorem of monotone convergence let $\epsilon$ there exist $n$ such that $\int_{E_{n}^c} |f|^p d\mu = \int |f|^p d\mu - \int _{E_n}|f|^p d\mu < \epsilon$ (remember that $\phi(E) = \int_E |f|^p d\mu$ is measure). – 2017-02-04

0 Answers 0