In the chapter you quote, Billingsley defines weak convergence of functions in $L^p$, which is by definition the convergence against every function in $L^q$, the dual of $L^p$. That is, $(f_n)$ in $L^p$ converges to $f$ in $L^p$ if and only if $\int f_ng$ converges to $\int fg$ for every $g$ in $L^q$ and this is a weak convergence because $L^q=(L^p)^*$.
In the chapter you quote, Chung defines weak convergence of functions in $L^1$, which is by definition the convergence against every function in $L^\infty$, the dual of $L^1$. That is, $(f_n)$ in $L^1$ converges to $f$ in $L^1$ if and only if $\int f_ng$ converges to $\int fg$ for every $g$ in $L^\infty$ and this is a weak convergence because $L^\infty=(L^1)^*$.
In the page you quote, Wikipedia defines weak convergence of (probability) measures. Actually, this mode of convergence should be called weak* rather than weak (but usage is to call it weak) because it refers to the convergence against any bounded continuous functions, and the space $M_1$ of probability measures is included in the dual of the space $C_b$ of bounded continuous functions. That is, $(\mu_n)$ in $M_1$ converges to $\mu$ in $M_1$ if and only if $\int f\mathrm d\mu_n$ converges to $\int f\mathrm d\mu$ for every $f$ in $C_b$ and this is a weak* convergence because $M_1\subset(C_b)^*$.
To mean that the distributions of some random variables converge weakly in the sense above, one usually says that the random variables converge in distribution.
There was such a question: "Is there anyway that these two types of convergence are related?"
Suppose we have $X_n \to X$ almost surely.
Put $P_n (A) = P(X_n \in A)$ and $P(A) = P(X \in A)$ for all Borel set $A$. Then $P_n$, $n \ge 1$ and $P$ are probability measures on real line.
$X_n$ converges to $X$ almost surely, hence $X_n$ converges to $X$ in distribution, and a sequence of measures $P_n$ converges weakly to the measure $P$. According to Portmanteau Theorem (see Billingsley, Convergence of probability measures, 1999, 2nd edition, pp.15-16) it follows that $P_n(A) \to P(A)$ for all Borel sets $A$ such that $P( \partial A) = 0$. The next example shows that condition $P_n(A) \to P(A)$ holds not for all Borel sets $A$.
Consider a probability space $(\Omega, \mathcal{F}, \mu)$ where $\Omega = [0,1]$, $\mathcal{F}$ is sigma-algebra of Borel sets of $\Omega$ and $\mu$ is a standard Lebesgue measure, that is $\mu([a,b]) = b-a$. Put $\xi_n(\omega) = n \omega $ for $\omega \in [0, \frac{1}n ]$ and $\xi_n(\omega) = 0$ otherwise. In this case $P_n \sim U[0, \frac{1}{n}]$, $\xi_n \to X \equiv 0$, $P$ is Dirac measure at zero.
For $A = [ - 1, 0] $ we have $P_n(A) = 0 \nrightarrow P(A) = 1$.
As a result we get the next statement. If $X_n$ converges to $X$ almost surely then for corresponding measures $P_n$ and $P$ we have $P_n(A) \to P(A)$ for all Borel sets $A$ such that $P( \partial A) = 0$ but not for all $A$.
Now suppose we have convergence $P_n(A) \to P(A)$ for all $A$ from some sigma-algebra. If these measures are not probability measures of if these measures are not measures on $\mathbb{R}$ with Borel sigma-algebra, then there is no direct relationship between them and any random variables. Suppose that $P_n(A)$, $P(A)$ are probability measures on $\mathbb{R}$ with Borel sigma-algebra and $P_n(A) \to P(A)$ for all Borel sets $A$. Hence $P_n$ converges to $P$ weakly (Portmanteau Theorem). According to the Skorokhod's representation theorem we can construct random variables $X_n$, $X$ such that $X_n$ has distribution $P_n$, $X$ has distribution $P$, $X_n \to X$ almost surely and moreover $X_n$ and $X$ are defined of probability space $(\Omega, \mathcal{F}, \mu)$, where $\Omega = [0,1]$, $\mathcal{F}$ consists of Borel sets of $\Omega$ and $\mu$ is a standard Lebesgue measure, that is $\mu([a,b]) = b-a$.
As a result we get the next statement. If $P_n(A)$, $P(A)$ are probability measures on $\mathbb{R}$ with Borel sigma-algebra and $P_n(A) \to P(A)$ for all Borel set $A$ then there are random variables $X_n, X$ on a "simple" probability space such that $X_n$ has distribution $P_n$, $X$ has distribution $P$ and $X_n \to X$ almost surely.
Best Answer
Weak convergence is weaker than convergence in measure. A very easy and illuminating example is to note that for weak convergence, the only requirement is that the distributions of the random variable converges. So if we pick for instance $X$ to be Gaussian (or something symmetric), and consider the sequence $X,-X,X,-X,\ldots$ (Ok, for your case you may need to pick a $[0,1]$ valued r.v. symmetric about $0.5$ and use $X,1-X,X,1-X,\ldots$). This sequence trivially converges weakly (as the distributions in the sequence are identical) but it's clear that this sequence does not converge in measure.
PS if the weak convergence is to a degenerate distribution (constant), then the convergence is also in measure, if I remember correctly.
If I interpret your remarks correctly, you are asking about the induced measures on $[0,1]$, throwing out the original sample space. Then you talk about the same notion of convergence, which is that $\mu_n(A) \to \mu(A)$ for sets $A$ without atoms of $\mu$, where these are measures on $[0,1]$. This notion relates to the weak* convergence of measures as linear functionals on the space of continuous functions. This is all in the realm of real analysis.
But then you ask about a sample space for probability and notions of convergence in measure (probability). For these, the sample space matters a lot! (As my example shows). To even talk about a sequence of random variables generically you have to use a sample space of the form $\Omega^\mathbb{N}$.
If we go back to just $[0,1]$, are you then asking how it relates to other notions of convergence? (ptwise, and in measure convergence make sense only for functions, say densities, but not for generic measures).