My answer here is realy just a footnote to Paul Siegel's excellent answer, but it has become too long to fit in a comment box. Integrals are siamese brothers to measures; leaving them out seems rather perverse to me. Anyway, here is how I think of integrals. The objective here is to tackle the "categorical" part; the analytical viewpoint will necessarily obtrude. But bear with me a little, this is a somewhat long post, with a punchline at the end.

Fix a Boolean algebra $\Omega$. A map $\nu: \Omega\to V$ with values on a linear space $V$ is finitely additive if $\nu(E\cup F)= \nu(E) + \nu(F)$ for every disjoint $E, F$. Denote the linear space of such maps by $\mathbf{A}(\Omega, V)$.

Theorem 1: There is a linear space $\mathbf{S}(\Omega)$ and a finitely additive map $\chi:\Omega\to \mathbf{S}(\Omega)$ universal among all finitely additive maps.
proof: just follow the universal property and do the obvious thing (yeah, I suppose you can use the adjoint functor theorem but why would you?).

The universal property recast in terms of representability gives the natural isomorphism ($\mathbf{Vect}$ is the category of linear spaces)

$$\mathbf{A}(\Omega, V)\cong \mathbf{Vect}(\mathbf{S}(\Omega), V)$$

Before continuing, let me elucidate a little bit of the structure of $\mathbf{S}(\Omega)$.

Theorem 2: Let $f$ be a non-zero element of $\mathbf{S}(\Omega)$. Then there are non-zero scalars $k_n$ and non-zero, pairwise disjoint $E_n\in \Omega$ such that $f= \sum_n k_n\chi(E_n)$. Furthermore, if $\nu$ is a finitely additive map and $\widehat{\nu}$ the map induced on $\mathbf{S}(\Omega)$ by universality, then $\widehat{\nu}(f)= \sum_{n}k_n \nu(E_n)$.

To put it simply, $\mathbf{S}(\Omega)$ is the linear space of "simple functions on $\Omega$" and the map induced by universality is the integral. Now use theorem 2 to put a norm on $\mathbf{S}(\Omega)$:

$$\|\sum_n k_n\chi(E_n)\|= \max\{|k_n|\}$$

Denote the completion by $\mathbf{L}_{\infty}(\Omega)$. On the other hand, put on the linear subspace of the *bounded* finitely additive maps $\Omega\to B$ with $B$ a Banach space, the semivariation norm (which I am not going to define). Denote this space by $\mathbf{BA}(\Omega, B)$. Then:

Theorem 3: There is a bounded finitely additive $\chi:\Omega\to \mathbf{L}_{\infty}(\Omega)$ universal among all bounded finitely additive maps.

Once again, recasting the universal property in terms of representability, we have a natural isometric isomorphism ($\mathbf{Ban}$ is the category of Banach spaces and bounded linear maps).

$$\mathbf{BA}(\Omega, B)\cong \mathbf{Ban}(\mathbf{L}_{\infty}(\Omega), B)$$

It is illuminating to write down what does the naturality of the isomorphism implies: I will leave that as an exercise to the reader.

Note that $\mathbf{L}_{\infty}(\Omega)$ is a Banach algebra in a natural way (use theorem 2 or juggle the universal property around. Or "cheat" all the way up and use Stone duality) and that $\chi$ is *spectral* or *multiplicative*, that is, $\chi(E\cap F)= \chi(E)\chi(F)$. Theorem 3 can now be extended by saying that $\chi$ is universal among all spectral measures (with values in Banach algebras). This extension is trivial given theorem 3.

The case of $\mathbf{L}_{\infty}(\Omega)$ does not need the introduction of measures but of course, this is not so with $\mathbf{L}_{1}$. So fix a finitely additive, positive $\mu:\Omega\to \mathbb{R}$. For the sake of simplification I will assume $\mu$ non-degenerate, that is, $\mu(E)= 0$ implies $E= 0$ (otherwise, you will have to take some quotient along the way). A finitely additive $\nu:\Omega\to B$ with $B$ a Banach space is $\mu$-Lipschitz if there is a constant $C$ such that $\|\nu(E)\|\leq C\mu(E)$ for all $E$. The infimum of all the constants $C$ in the conditions of the inequality gives a norm and a normed space I will denote by $\mathbf{LA}(\Omega, \mu, B)$. On the other hand, endow $\mathbf{S}(\Omega)$ with the norm

$$\|\sum_n k_n\chi(E_n)\|= \sum_n |k_n|\mu(E_n)$$

and denote the completion by $\mathbf{L}_{1}(\Omega, \mu)$.

Theorem 4: There is a finitely additive, $\mu$-Lipschitz $\chi:\Omega\to \mathbf{L}_{1}(\Omega, \mu)$ universal among all such maps.

Before the conclusion let me address a few points.

Measurable spaces are not needed. If you really want them, use Stone duality (that is, points count for nothing in measure theory so why not leave them out, heh?).

Finitely additive measures are really not that much more general than $\sigma$-additive ones. I will leave this cryptic comment as is, and just note that once again, Stone duality is the key here.

I am *not* advocating this approach to be used in teaching (unless your goal is to flunk and befuddle as many undergrads as humanly possible). For one, you need some functional analysis under the belt (Banach spaces, completions, semivariation, etc.). Intuition is very hard to come by as I have thrown away the measurable spaces without which THE most important example, Lebesgue measure (arguably, the core of a first measure theory course) cannot be constructed. The whole logic of the approach only makes sense after you have seen other instances of categorical thinking at work. I am sure you can think of other objections.

How categorical is this approach? Certainly, the universal properties of the respective spaces are central to the whole business and at least, they make clear that some results are really just a consequence of abstract nonsense. In the words of P. Freyd, category theory is doing what it was invented for: to make the easy things really easy (or some such, my memory is lousy). For example, the Bochner vector integral is obtained simply by taking the projective tensor product. Fubini and Fubini-Tonelli on the equality of iterated integrals are other notable cases of categorical thinking at work. Now pepper with Stone duality and a few more tools (e.g. Hahn-Banach and the compact-Hausdorff monad) and you can get (a slight variation of) the Riesz representation theorem for compact Hausdorff spaces. Use the proper compactifications and generalize to wider classes of topological spaces. Or use Loomis-Sikorski to get Vitali-Hahn-Saks in one line (but this is really "cheating" as the crucial step in establishing Loomis-Sikorski is essentially the same as the one to establish Vitali-Hahn-Saks: a Baire category-theorem application). And a few more.

But once again, how categorical is this approach? Well, the argument is categorical enough to be generalized to symmetric monoidal closed categories. See R. Borger -- A categorical approach to integration, in the 23rd volume of TAC available online. For the modifications needed to internalize the arguments to a topos (and much more) see the delightful Phd thesis of Mathew Jackson "A Sheaf theoretic approach to measure theory" -- this is available online, just google for it. Oh, by the way, you can see (almost) everything I have explained above in volume 3 of D. Fremlin's measure theory 5-volume series, also available online.