Measurable function

**slevvio** · April 28th 2011, 05:34 PM

Hello everyone I was wondering if you could help me with something. Let $f=\displaystyle\sum_{n=1}^{\infty}4^n \chi_{E_n}$ for $x \in \mathbb{R}$ where $E_n = (0,2^{-n}]$ and chi is the characteristic function. Then f is measurable since it is the limit pointwise (almost) everywhere of step functions, but how do I show that $\int f = \infty$ ?

If I define $f_n = \displaystyle\sum_{r=1}^n 4^n \chi_{E_n}$ , then of course

$\int f_n = \displaystyle\sum_{r=1}^n 2^r$ .

But since the limit of this sequence is infinity, does this tell me that the integral of f diverges? Can I say that the integral of a measurable function $\int f = \lim_{n\rightarrow\infty} \int f_n$ ?

The definition of the integral of a non-negative measurable function is that you take the truncation $f(x)=n ($ if $f(x)\ge n) , -n$ (if $f(x) \le -n$ ) and leave it the same otherwise, then
$\int f := \lim_{n \rightarrow \infty} \chi_{[-n,n]}f_n$

so in the context of my question I guess I am trying to show that

$\int f =\lim_{m\rightarrow \infty} \int \chi_{[-m,m]}\left(\displaystyle\sum_{n=1}^{\infty}4^n\chi_{E_ n}\right)_m=\infty$ ,

but that seems like a giant mess I can't make sense of. Can anyone give me a hand with this. Thanks a lot

**Opalg** · April 29th 2011, 12:12 AM

Originally Posted by slevvio

Can I say that the integral of a measurable function $\int f = \lim_{n\rightarrow\infty} \int f_n$ ?

In general, no. But the famous convergence theorems of Lebesgue give you some conditions under which the answer is yes. The one that you want here is Lebesgue's monotone convergence theorem.

**slevvio** · April 29th 2011, 01:29 AM

We studied that only for functions in L^1(R).

So does the result hold when the sequence of integrals diverges?

**girdav** · April 29th 2011, 03:38 AM

The result holds true because $\int fd\lambda \geq \int f_nd\lambda$ for all $n$ .

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