I will concentrate on comparing (3) and (4). The definition (1) is meant for finite signed measures, whereas all the other definitions are meant for arbitrary positive measures; (1) is equivalent to (4) in the case of finite positive measures. (2) appears to be equivalent to (4) ["locally finite" can mean "finite on compact sets", although it is sometimes taken to mean "finite on the elements of some topological basis"; these are equivalent in the LCH (locally compact Hausdorff) case]. Finally, (5) does not appear to be a definition at all, but rather a description of a definition.
Now then,
i) In the case of a second countable LCH space, every locally finite measure satisfies both (3) and (4) (Theorem 7.8 of [1]). This is the most commonly considered scenario in applications, which is why almost no one bothers to carefully sort out the differences between the different definitions.
ii) In the case of a sigma-compact LCH space, (3) and (4) are equivalent. The forward direction is Corollary 7.6 of [1]; the backwards direction follows from the forward direction together with (iv) below (but I'm sure there is an easier proof).
iii) (3) and (4) are not equivalent in general, even for LCH metrizable spaces (Exercise 7.12 of [1]).
iv) In an LCH space, there is a bijection between
A) measures satisfying (3),
B) measures satisfying (4), and
C) positive linear functionals on the space of continuous functions with compact support.
(The Riesz representation theorem gives either (A)<->(C) or (B)<->(C), depending on where you look; (A)<->(B) is in the Schwarz book mentioned by Joe Lucke; see also Exercise 7.14 of [1])
[1] G. B. Folland, Real Analysis: Modern Techniques and Their Applications
Note: In [1], "Radon" refers to measures satisfying (3).
First, we will see how Billingsley's claim helps to conclude. For each $n$, consider $K_n$ compact such that $\mu(K_n)>1-n^{-1}$. We can assume the sequence $(K_n,n\geqslant 1)$ to be increasing. Take $B\in\mathcal B(X)$ and $\varepsilon>0$. By the claim, there is $F\subset B$ such that $\mu(B\setminus F)<\varepsilon$. Fix $n$ such that $\varepsilon>n^{-1}$. Then $$\mu(B\cap K_n)=\mu(B)-\mu(B\setminus (B\cap K_n))=\mu(B)-\mu(B\setminus K_n)\\\geqslant \mu(B)-\mu(K_n^c)\geqslant \mu(B)-n^{-1}\geqslant \mu(B)-\varepsilon.$$
We thus get $\mu(F\cap K_n)>\mu(B)-2\varepsilon$, and $F\cap K_n$ is a compact subset of $B$.
Now we shall see the claim is true. First, note that in a metric space, an closed set is a countable intersection of open sets, and using the fact that the measure is finite, we deduce that the assertion is true for $B$ open.
Then the remaining task is to show that the collection of the $B$ for which the property holds is a $\sigma$-algebra.
Best Answer
Wikipedia is wrong because its definition of a tight measure as "for all ε>0, there is some compact subset K of X such that μ(X−K)<ε" fails to note that the measure must be a probability (or totally finite) measure. (Parthasarathy, 1967, Probability Measures on Metric Spaces, p. 26, fn 1.) Otherwise, a measure μ is said to be tight, if for every set A∈Σ, μ(A)=sup{μ(K)| compact K⊆A} and inner regular, if for every set A∈Σ, μ(A)=sup{μ(F)| F closed and F⊆A}.