Breaking up domain of integration for a Lebesgue Integral I am hesitant in all my pronouncements of what one can do with Lebesgue integrals, since I am just now figuring them out.
For example, it seems clear one can split up in this way:
$$\int_{[0,1]}f d\mu = \int_{[0,1\frac{1}{2}]} f d\mu + \int_{[\frac{1}{2},1]} f d\mu,$$
but can one do this $$\int_{\mathbb{R}^+}f d\mu = \int_{[0,1]} f d\mu + \int_{[1,2]} f d\mu + \dots$$ so long each of the integrals over $[n,n+1]$ is well defined?
In particular, say $$\int_{[n,n+1]} f d\mu \ge \frac{1}{n}$$ I would like to argue that $\int_{\mathbb{R}^+}f d\mu = \infty$ by splitting it up into these bits that, in sum, diverge.
I guess an argument goes something like: $$\int_{\mathbb{R}^+}f d\mu \ge \int_{[0,1]} f d\mu + \int_{[1,2]} f d\mu + \dots\int_{[n,n+1]} f d\mu \ge \sum^n \frac{1}{n} $$ and then 'take a limit'?
But I know there are some complications when it comes to trying to treat $\int_{\mathbb{R}^+}f d\mu$ as an improper integral in the Riemann sense. Are these complications relevant to the argument I am trying to give above.
Any general advice about how to think in this terrain would be great. Thanks!
 A: First of all: it depends whether you are dealing with non-negative functions or any functions.
With non negative functions, you can do almost anything you want. Sometimes you get your integral equal to infinity, but for example your splits into intervals $[n,n+1]$ do work just like you described it.
It's when you have to integrate a generic function, when things become more complicated. Concerning your splits, it is not enough for the integral to be well defined on each $[n,n+1]$. For example the function sin is integrable on every compact, but you can't integrate it over $\mathbb{R}^+$.
If your function $f$ however is integrable on $\mathbb{R}^+$, then you can split Here is a proof:
We have $\int \limits_{0}^n f(t)dt=\int \limits_{\mathbb{R}^+} f \mathbb{1}_{[0,n]}$. Pointwise, the sequence of functions $f \mathbb{1}_{[0,n]}$ converges to $f$ and they are all dominated in absolute value by $|f|$, that is integrable. Therefore by the theorem of dominated convergence, the integral $\int \limits_{0}^n f(t)dt$ converges to $\int \limits_{\mathbb{R}^+}f$.
