An example of convergence to Young measures $\newcommand{\R}{\mathbb{R}}$
$\newcommand{\lam}{\lambda}$
I am trying to prove the following claim:
Let $\{u:[0,1]\to \mathbb{R} \mid u \, \, \text{  is differentiable a.e}, u(0)=u(1)=0 \}^{*} $. Consider the functional $I(u)=\int_0^1 ((u')^2-1)^2+u $ 
For every minimizing sequence $u_n$ of $I$, the sequence of derivatives $u_n'$ generates the Young measures $\nu_x =\frac{1}{2}(\delta_1+\delta_{-1})$ 
(This is taken from wikipedia).
So far I have succeeded in proving** this only for a specific minimizing sequence, which is a natural one to consider. My proof uses the specific structure of this sequence, so it is not applicable as is to the general case.
My question then, is how to prove this is true for every minimizing sequence?

(*) Perhaps the right space should be some Sobolev space (and then the boundary condition is in the trace sense).
(**) Here is my proof:
Define $A_1=[0,\frac{1}{2}],A_2=[0,\frac{1}{4}] \cup[\frac{2}{4},\frac{3}{4}],A_3=[0,\frac{1}{8}]\cup [\frac{2}{8},\frac{3}{8}]\cup [\frac{4}{8},\frac{5}{8}]\cup [\frac{6}{8},\frac{7}{8}]$, and so on for $n \in \mathbb{N}$. Define a sequence of ``zig-zag triangle'' functions, such that $|u_n| \le \frac{1}{2^n}$, and 
$$ u'_n(x) =
\begin{cases}   
1,  & \text{if $x\in A_n$} \\
-1, & \text{if $x \notin A_n$}
\end{cases} 
$$
Then $(u'_n)^2=1$ a.e so $I(u_n) \le \frac{1}{2^{2n}} \rightarrow 0$.
Let $f:\R \to \R$ be a continuous function satisfying  $\lim_{|\lambda|\to\infty}f(\lambda) =0$. Then
$$ (f \circ u'_n)(x) =
\begin{cases}   
f(1),  & \text{if $u_n'(x) =1$} \\
f(-1), & \text{if $u_n'(x) = -1$}
\end{cases} 
$$
i.e, 
$$ (f \circ u'_n)(x) =
\begin{cases}   
f(1),  & \text{if $x\in A_n$} \\
f(-1), & \text{if $x \in [0,1] \setminus A_n$}
\end{cases} 
$$
So, denoting $f \circ u_n' =F_n$, we need to show $F_n$ converges in the weak* topology on $L^\infty([0,1])$ to the function:
$F(x) = \int_{\R} f d\nu_x =\frac{1}{2}(f(1)+f(-1))$.
By definition of weak* convergence in $L^{\infty}=\left( L^1 \right)^*$ this means that for every $g \in L^{1}([0,1])$ :
$$(1) \lim_{n \to \infty} \int_0^1 F_n(x)g(x)\,dx= \int_0^1 F(x)g(x)\,dx= \frac{1}{2}(f(1)+f(-1)) \cdot  \int_0^1 g(x)\,dx $$
Now,
$$ \int_0^1 F_n(x)g(x)\,dx= f(1) \int_{A_n} g(x)\,dx + f(-1) \int_{[0,1] \setminus A_n} g(x)\,dx$$
Noting that $[0,1] \setminus A_n = A_n + \frac{1}{2^n}$, we get 
$$(2) \int_0^1 F_n(x)g(x) \, dx= f(1) \int_{A_n} g(x) \, dx + f(-1) \int_{A_n + \frac{1}{2^n}} g(x)\,dx$$
So, by $(1),(2)$ it's enough to prove $$ (3) \lim_{n \to \infty} \int_{A_n} g(x)\,dx = \lim_{n \to \infty}  \int_{A_n+\frac{1}{2^n}} g(x) \, dx= \frac{1}{2}  \int_0^1 g(x)\,dx $$
We turn to handle the second integral:
First, note that we can extend $g$ to $\R$ by putting $g|_{\R \setminus \left[ 0,1 \right]}=0$.
(Now we think of $g \in L^1(\R)$). 
By variables changing: $y=x-\frac{1}{2^n}$ we get:
$$(4) \int_{A_n + \frac{1}{2^n}} g(x)dx = \int_{A_n}g\left(y+\frac{1}{2^{n}}\right) \,dy=\int_{A_n} g_{\frac{1}{2^n}}(y)dy $$
where $g_h(y)=g(y+h)$ (This is part of the reason we needed to extend $g$ to all $\R$, so now we do not need to worry about what happens when we ``translate'' outside the limits where the original $g$ is defined).
By the $L^1$-continuity of translation,we know:
$$ \lim_{h \to 0} \int_\R |g_h(x)-g(x)| \, dx =0, $$ so in particular:
$$ \lim_{n \to \infty} \int_\R |g_{\frac{1}{2^n}}(x)-g(x)|\,dx = 0, $$
and since $$ \int_\R \left|g_{\frac{1}{2^n}}(x)-g(x)\right|\,dx \ge  \int_{A_n} \left|g(x)_{\frac{1}{2^n}}-g(x)\right|\,dx \ge 0 $$
we get:
$$ \lim_{n \to \infty} \int_{A_n} \left|g_{\frac{1}{2^n}}(x)-g(x)\right|\,dx = 0, $$
so
$$ \lim_{n \to \infty} \left| \int_{A_n} g_{\frac{1}{2^{n}}}(x)-g(x)\,dx \right| = 0, $$
which by $(4)$ is equivalent to:
$$ \lim_{n \to \infty} | \int_{A_n+\frac{1}{2^{n}}} g(x) -\int_{A_n}g(x)dx \,  | = 0, $$
However, it is also holds that:
$$ \int_{A_n+\frac{1}{2^{n}}} g(x) + \int_{A_n}g(x)dx = \int_0^1 g(x)dx$$
So, after denoting $a_n = \int_{A_n+\frac{1}{2^{n}}} g(x)dx \, , \, b_n =\int_{A_n}g(x)dx, c=\int_0^1 g(x)dx $, we are in the following situation:
$$a_n+b_n=c, \lim_{n \to \infty}|a_n-b_n|=0 $$
This implies: $2a_n=(a_n+b_n)+(a_n-b_n)=c +(a_n-b_n) \Rightarrow \lim_{n \to \infty} a_n =\frac{c}{2}  $, so actually
$ \lim_{n \to \infty} b_n =\lim_{n \to \infty} a_n =\frac{c}{2}  $
In our context, this translates into:
$$ \lim_{n \to \infty} \int_{A_n} g(x)\,dx = \lim_{n \to \infty}  \int_{A_n+\frac{1}{2^{n}}} g(x)\,dx= \frac{1}{2} \int_0^1 g(x)\,dx $$
Which is exactly $(3)$ we wanted to show.
 A: I suggest you to read the lectures notes by Müller which are linked in the Wikipedia article, I think they offer a nice introduction to the topic, and they solve this problem as Example b) in section 3.2. To sum up the argument:
0) The right space is indeed a Sobolev space, e.g. $X=\{u\in W^{1,1}(0,1): u(0)=u(1)=0\}$ where the boundary values are in the trace sense, or you could use the fact that in dimension $1$ Sobolev functions are uniformly continuous therefore you can define them pointwise.
1) You can prove that $\inf_{u\in X} I(u)=0$  using the same sequence you already used. Consider then a minimizing sequence $(u_n)$ so that $I(u_n)\to 0$. In particular the sequence $(\dot u_n)$ is bounded in $L^4$, and there is a subsequence $(\dot u_{n_k})$ which generates probability Young measures $\nu_x$. Moreover the sequence $(u_n)$ is bounded in $W^{1,4}$ and $u_n\to 0$ in $L^2$, therefore we can also suppose that the subsequence $\dot u_{n_k}$ converges weakly to zero (for instance $W^{1,\infty}$-weakly*).
2) Set $g(p)=\min\{(p^2-1)^2,1\}$. This is a bounded continuous function, and from the boundedness in $L^4$ of $\dot u_{n_k}$ in the limit we obtain(*)
$$\int\limits_0^1dx\int\limits_{\mathbb R}g(p)d\nu_x(p)=\lim_{k\to\infty} \int\limits_0^1 g(\dot u_{n_k}(x))dx\leq\lim_{k\to\infty} I(u_{n_k})=0.$$
Therefore  for a.e. $x\in(0,1)$ 
$$\int\limits_{\mathbb R}g(p)d\nu_x(p)=0$$
which implies that $\mathrm{supp}\,\nu_x\subset\{0,1\}$. We thus obtain $\nu_x=\lambda(x) \delta_1+(1-\lambda(x))\delta_{-1}$.
3) To show that $\lambda(x)=\frac12$ for a.e. $x$, use the limit properties of Young measures against the function $g(p)=p$, and obtain(*)
$$\dot u_{n_k}{\rightharpoonup}\int\limits_{\mathbb R} p \,d\nu_x(p)=\lambda(x)-(1-\lambda(x))=2\lambda(x)-1$$
which together with the fact that $\dot u_{n_k}\rightharpoonup 0$ implies $\lambda(x)=\frac12$ for a.e. $x$.
4) Since the Young measure thus obtained from the subsequence is uniquely determined, it follows that the full sequence generates the same Young measure, i.e. $\frac12\delta_1+\frac12\delta_{-1}$.
(*) Note: you have to be careful and check that you can pass to the limit when you take the composition with the function $g$: in general you can do it only for functions in $C_0(\mathbb R)$, but if you have e.g. some bound in $L^p$, $p>1$ for the sequence which generates the Young measures, then you can take $g$ with linear growth (thus $g(p)=p$ is ok).
