Measurability

Sarah · **Posted:** Mon, 23 Mar 2009 10:52:54 UTC

Let $(X,A, \mu)$ be a measurable space with $\mu(X) < 1$ , and let $f : (X,A) \rightarrow (R, B)$ be
an integrable positive function (B is Borel-set), i.e. satisfying
$\int_X fd\mu < \infty$
Show that the integral can be calculated as the limit
$\int_X fd\mu = \lim_{n\rightarrow\infty} \sum_{k=0}^{\infty}z_{n,k} \mu\{x\in X : y_{n,k} \leq f(x) < y_{n,k+1}\}$
where for every n\in N
$0 < y_{n,0} < y_{n,1} < y_{n,2} < \ldots, \lim_{k\rightarrow\infty}y_{n,k} = \infty$
$z_{n,k}$ is any number satisfying $y_{n,k} \leq z_{n,k} \leq y_{n,k+1}$ and
$\lim_{n\rightarrow\infty}\sup_{k\geq0}\left(y_{n,k+1}-y_{n,k}\right)$

I actually understand the question, I just have no idea how to do it, how to write it down or work it out.

outermeasure · **Posted:** Mon, 23 Mar 2009 14:41:04 UTC

Sarah wrote:

Let $(X,A, \mu)$ be a measurable space with $\mu(X) < 1$ , and let $f : (X,A) \rightarrow (R, B)$ be
an integrable positive function (B is Borel-set), i.e. satisfying
$\int_X fd\mu < \infty$
Show that the integral can be calculated as the limit
$\int_X fd\mu = \lim_{n\rightarrow\infty} \sum_{k=0}^{\infty}z_{n,k} \mu\{x\in X : y_{n,k} \leq f(x) < y_{n,k+1}\}$
where for every n\in N
$0 < y_{n,0} < y_{n,1} < y_{n,2} < \ldots, \lim_{k\rightarrow\infty}y_{n,k} = \infty$
$z_{n,k}$ is any number satisfying $y_{n,k} \leq z_{n,k} \leq y_{n,k+1}$ and
$\lim_{n\rightarrow\infty}\sup_{k\geq0}\left(y_{n,k+1}-y_{n,k}\right)$

I actually understand the question, I just have no idea how to do it, how to write it down or work it out.

I think your last condition should be
$\displaystyle \lim_{n\to\infty}\sup_{k\geq 0}\left(y_{n,k+1}-y_{n,k}\right)=0.$
If f is a simple function the result holds. So approximate f by simple functions ...

Opalg · **Posted:** Mon, 23 Mar 2009 14:41:44 UTC

Sarah wrote:

Let $(X,A, \mu)$ be a measurable space with $\mu(X) < 1$ , and let $f : (X,A) \rightarrow (R, B)$ be
an integrable positive function (B is Borel-set), i.e. satisfying
$\int_X fd\mu < \infty$
Show that the integral can be calculated as the limit
$\int_X fd\mu = \lim_{n\rightarrow\infty} \sum_{k=0}^{\infty}z_{n,k} \mu\{x\in X : y_{n,k} \leq f(x) < y_{n,k+1}\}$
where for every n\in N
$0 < y_{n,0} < y_{n,1} < y_{n,2} < \ldots, \lim_{k\rightarrow\infty}y_{n,k} = \infty$
$z_{n,k}$ is any number satisfying $y_{n,k} \leq z_{n,k} \leq y_{n,k+1}$ and
$\lim_{n\rightarrow\infty}\sup_{k\geq0}\left(y_{n,k+1}-y_{n,k}\right)$

I actually understand the question, I just have no idea how to do it, how to write it down or work it out.

Let $g_n(x) = \sum_{k=0}^{\infty}z_{n,k}\chi_{n,k}(x)$ , where $\chi_{n,k}(x)$ is the function that takes the value 1 on the set $\{x\in X : y_{n,k} \leq f(x) < y_{n,k+1}\}$ and is zero elsewhere. Notice that g_k

is integrable, with integral $\sum_{k=0}^{\infty}z_{n,k} \mu\{x\in X : y_{n,k} \leq f(x) < y_{n,k+1}\}$ , and that $|g_k(x) - f(x)|\leqslant \sup_{k\geq0}\left(y_{n,k+1}-y_{n,k}\right)$ for all x in X. Thus $\Bigl|\int_Xf\,d\mu - \int_Xg_n\,d\mu\Bigr|\leqslant \mu(X)\sup_{k\geq0}\left(y_{n,k+1}-y_{n,k}\right)$ .

Sarah · **Posted:** Mon, 23 Mar 2009 21:36:03 UTC

Of course the last condition is incorrect, thanks for noticing!
The help looks great, I going to write it down completely...

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Measurability

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