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