The Context
(In this section I'm just going to explain hypothesis testing, type one and two errors, etc, in my own style. If you're comfortable with this material, skip to the next section)
The Neyman-Pearson lemma comes up in the problem of simple hypothesis testing. We have two different probability distributions on a common space $\Omega$: $P_0$ and $P_1$, called the null and the alternative hypotheses. Based on a single observation $\omega\in\Omega$, we have to come up with a guess for which of the two probability distributions is in effect. A test is therefore a function which to each $\omega$ assigns a guess of either "null hypothesis" or "alternative hypothesis". A test can obviously be identified with the region on which it returns "alternative", so we're just looking for subsets (events) of the probability space.
Typically in applications, the null hypothesis corresponds to some kind of status quo, whereas the alternative hypothesis is some new phenomenon which you're trying to prove or disprove is real. For example, you may be testing someone for psychic powers. You run the standard test with the cards with squiggly lines or what not, and get them to guess a certain number of times. The null hypothesis is that they'll get no more than one in five right (since there's five cards), the alternative hypothesis is that they're psychic and may get more right.
What we'd like to do is minimize the probability of making a mistake. Unfortunately, that's a meaningless notion. There are two ways you could make a mistake. Either the null hypothesis is true, and you sample an $\omega$ in your test's "alternative" region, or the alternative hypothesis is true, and you sample the "null" region. Now, if you fix a region $A$ of the probability space (a test), then the numbers $P_0(A)$ and $P_1(A^{c})$, the probabilities of making those two kinds of errors, are completely well-defined, but since you have no prior notion of "probability that the null/alternative hypothesis is true", you can't get a meaningful "probability of either kind of mistake". So this is a fairly typical situation in mathematics where we want the "best" of some class of objects, but when you look closely, there is no "best". In fact, what we're trying to do is minimize $P_0(A)$ while maximizing $P_1(A)$, which are clearly opposing goals.
Keeping in mind the example of the psychic abilities test, I like to refer to the type of mistake in which the null is true but you conclude the alternative as true as "delusion" (you believe the guy's psychic but he's not), and the other kind of mistake as "obliviousness".
The Lemma
The approach of the Neyman-Pearson lemma is the following: let's just pick some maximal probability of delusion $\alpha$ that we're willing to tolerate, and then find the test that has minimal probability of obliviousness while satisfying that upper bound. The result is that such tests always have the form of a likelihood-ratio test:
Proposition (Neyman-Pearson lemma)
If $L_0, L_1$ are the likelihood functions (PDFs) of the null and alternative hypotheses, and $\alpha > 0$, then the region $A\subseteq \Omega$ which maximizes $P_1(A)$ while maintaining $P_0(A)\leq \alpha$ is of the form
$$A=\{\omega\in \Omega \mid \frac{L_1(\omega)}{L_0(\omega)} \geq K \}$$
for some constant $K>0$. Conversely, for any $K$, the above test has $P_1(A)\geq P_1(B)$ for any $B$ such that $P_0(B)\leq P_0(A)$.
Thus, all we have to do is find the constant $K$ such that $P_0(A)=\alpha$.
The proof on Wikipedia at time of writing is a pretty typically oracular mathematical proof that just consists in conjecturing that form and then verifying that it is indeed optimal. Of course the real mystery is where did this idea of taking a ratio of the likelihoods even came from, and the answer is: the likelihood ratio is simply the density of $P_1$ with respect to $P_0$.
If you've learned probability via the modern approach with Lebesgue integrals and what not, then you know that under fairly unrestrictive conditions, it's always possible to express one probability measure as being given by a density function with respect to another. In the conditions of the Neyman-Pearson lemma, we have two probability measures $P_0$, $P_1$ which both have densities with respect to some underlying measure, usually the counting measure on a discrete space, or the Lebesgue measure on $\mathbb R^n$. It turns out that since the quantity that we're interested in controlling is $P_0(A)$, we should be taking $P_0$ as our underlying measure, and viewing $P_1$ in terms of how it relates to $P_0$, thus, we consider $P_1$ to be given by a density function with respect to $P_0$.
Buying land
The heart of the lemma is therefore the following:
Let $\mu$ be a measure on some space $\Omega$, and let $f$ be a positive, integrable function on $\Omega$. Let $\alpha > 0$. Then the set $A$ with $\mu(A)\leq\alpha$ which maximizes $\int_A fd\mu$ is of the form
$$\{\omega\in\Omega\mid f(\omega)\geq K\}$$
for some constant $K>0$, and conversely, any such set maximizes $\int f$ over all sets $B$ smaller than itself in measure.
Suppose you're buying land. You can only afford $\alpha$ acres, but there's a utility function $f$ over the land, quantifying, say, potential for growing crops, and so you want a region maximizing $\int f$. Then the above proposition says that your best bet is to basically order the land from most useful to least useful, and buy it up in order of best to worst until you reach the maximum area $\alpha$. In hypothesis testing, $\mu$ is $P_0$, and $f$ is the density of $P_1$ with respect to $P_0$ (which, as already stated, is $L_1/L_0$).
Here's a quick heuristic proof: out of a given region of land $A$, consider some small one meter by one meter square tile, $B$. If you can find another tile $B'$ of the same area somewhere outside of $A$, but such that the utility of $B'$ is greater than that of $B$, then clearly $A$ is not optimal, since it could be improved by swapping $B$ for $B'$. Thus an optimal region must be "closed upwards", meaning if $x\in A$ and $f(y)>f(x)$, then $y$ must be in $A$, otherwise we could do better by swapping $x$ and $y$. This is equivalent to saying that $A$ is simply $f^{-1}([K, +\infty))$ for some $K$.
