Thread: Measurable function

Hello everyone I was wondering if you could help me with something. Let $\displaystyle f=\displaystyle\sum_{n=1}^{\infty}4^n \chi_{E_n}$ for $\displaystyle x \in \mathbb{R}$ where $\displaystyle E_n = (0,2^{-n}]$ 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 $\displaystyle \int f = \infty$ ?

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

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

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 $\displaystyle \int f = \lim_{n\rightarrow\infty} \int f_n$?

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

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

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

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

Originally Posted by slevvio

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

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.

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

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

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