I'm working on two problems where I need to interchange the limit and integral signs, so I want to evoke Lebesgue's Dominated Convergence Theorem. I now know that the functions I have chosen do indeed dominate, but is there a way to show that the dominating functions are in fact Lebesgue integrable? (I don't want to calculate their integrals...)
One of the functions I am trying to show are Lebesgue integrable are: $\dfrac{1} {1+x^2}$ over the domain $[0,\infty)$
How would I go about doing this?
Since you asked for more "theoretical" approach, maybe this is something for you.
We have \begin{equation} \lim_{x \rightarrow \pm \infty}\frac{x^2}{1+x^2}=1 \end{equation} and therefore \begin{equation} \frac{1}{1+x^2} \leq c \ \frac{1}{x^2}, \end{equation} for $|x| \geq a$ and suitable constants $a$ and $c$. On the (compact) interval $ [ -a,a ]$ we can bound the (continuous) function by a constant. And outside of the interval it is just $\frac{1}{x^2}$ scaled by a constant which is known to be integrable outside an interval around zero.
In this case it is quite easy to see what $c$ and $a$ are (even without the limit argument, as pointed out by David C. Ullrich) - but this approach may work in "more complicated" cases: The function considered is continuous and behaves like a integrable function when $x \rightarrow \pm \infty$.
In this case, the evaluation of the improper integral is just an application of the Monotone Convergence Theorem to $f_n(x)=\frac{1}{1+x^{2}}\chi _{[0,n)}(x)$
It's impossible to answer the general question you ask; too general.
For the specific example of $f(x)=\frac{1}{1+x^2}$ being in $L^1([0,\infty))$ you could give a detailed proof as below. Note that what's below is much more detail than anyone would actually write, except in a first-semester measure theory class. Also note that I'm not going to use that $\arctan$ thing; I'm not going to calculate the integral exactly. What I do below is better because the sort of thing I do below would also let you show that a function like $1/(1+x^6)$ was Lebesgue integrable, when you don't know an explicit antiderivative. So:
First, $f$ is measurable since it's continuous.
Now $$\int_0^1 f\le\int_0^1 1=1.$$
For every $A$ with $1<A<\infty$ the theorem that the Lebesgue integral of a continuous function on $[a,b]$ equals its Riemann integral shows that $$\int_1^A f\le\int_1^A\frac{dt}{t^2}=1-A^{-1}\le 1.$$So the Monotone Convergence Theorem shows that $$\int_1^\infty f\le 1.$$
(You should think about how I'm applying MCT here; what increasing sequence of positive functions am I talking about?)
So $\int_0^\infty|f|=\int_0^\infty f\le 1+1<\infty$.
If we're talking about $L^1(\mathbb{R})$, then we should show that the integral is finite (it is).
Also, recall that this neat function is also Riemann integrable - that means that it is Lebesgue integrable. Check this out.
