Understanding this integral from a measure theory perspective. I know that 
$$\int_1^\infty \frac{1}{x} dx$$ does not converge in regular calculus.  But I'm looking at $L^p$ spaces now and this integral is a good counter example for some things, but what is the measure theory reason that the integral doesn't converge?
 A: Since your question is not very precisely formulated, I will interpret it as follows: Is there a (purely) measure theoretic proof of $\int_1^\infty 1/x \,dx =\infty$, which in particular avoids the use of antiderivatives.
Indeed, there is. Assume towards a contradiction that $\int_1^\infty 1/x\,dx <\infty$. Define 
$$
f_n = \frac{1}{n} 1_{(1,n)}.
$$
I leave it to you to verify $0\leq f_n (x)\leq 1/x$ for all $x\in (1,\infty)$ and that $f_n(x)\to 0$ pointwise. Thus, by dominate convergence, we get
$$
\frac{n-1}{n} = \int_1^\infty f_n \,dx \to \int_1^\infty 0\,dx=0,
$$
a contradiction, since $(n-1)/n \to 1\neq 0$.
A: I feel like you're messing things up.
Measure theory helps you decide which functions you can integrate. 
Let $(X, \mathcal{A})$ be a measured space, and  $\mathcal{B}(\mathbb{R})$  (resp. $\mathcal{B}(\mathbb{R}^+)$) the Borel algebra over the (resp. positive) reals. 
We note $\mathcal{F}$ the set of measurable functions $(X, \mathcal{A}) \rightarrow (\mathbb{R}, \mathcal{B}(\mathbb{R}))$, $\mathcal{F}^+$ the set of measurable functions $(X, \mathcal{A}) \rightarrow (\mathbb{R}^+,\mathcal{B}(\mathbb{R}^+))$,
For $f \in \mathcal{F}^+$, you can consider the quantity $\int_X f(x)dx$, which has a value in $\mathbb{R}^+ \cup \{\infty\}$, also written : $\int : \mathcal{F^+} \rightarrow \mathcal{R}^+\cup \{+ \infty \}$
(If $f$ is not measurable, the integral is not defined. $\int_X f(x)dx$ does not exist).
This definition is then extended to $\mathcal{F}^+$  by considering the positive and negative parts of functions, and you have : $\int : \mathcal{F} \rightarrow \mathcal{R}\cup \{+ \infty, - \infty \}$

Once we have this defintion, we are able to define various functions (that will become norms) over the set of measurable functions : the $p$-norms, $|| \cdot ||_p : \mathcal{F} \rightarrow \mathcal{R}^+\cup \{+ \infty \}$
These $p$-norms have remarquable properties, and would be norms, if only they would not sometimes go up to infinity. So to avoid this, we define the spaces $\mathcal{L}_p \subset \mathcal{F}$, such that $|| \cdot ||_p$ is a well defined norm over $\mathcal{L}_p$, by removing the functions whose $p$-norm is infinity.

Finally, you define $L_p$ by quotienting $\mathcal{L}_p$.

Edit : so, back to your question, it is not because a function doesn't have a finite integral that the $p$-norm of this function isn't correctly defined.
