My argument (briefly):

Using density of the simple functions in $textit{L}^{infty}((0,1))$, i must prove the definition for weak converges holds, but i can restrict the argument to simple functions.

So given a simple function $overline{g}$ i know there exist costants $c_1=0<c_2<cdots<c_n=1$ such that $overline{g}$ is costant on these, so you can come to this partial conclusion:

$$lim_{nto +infty} int_{(0,1)}overline{g}u_ndm=lim_{nto +infty}sum_{j=1}^{k-1}overline{g}(c_i)int_{c_j}^{c_{j+1}}u_n(z)dz=sum_{j=1}^{k-1}overline{g}(c_j)lim_{nto +infty}int_{c_j}^{c_{j+1}}u_n(z)dz=$$

$$= sum_{j=1}^{k-1}overline{g}(c_j)biggl( lim_{nto +infty}int_{(0,1)}achi_{E_n^{(j)}}(z)dz + lim_{nto +infty}int _{(0,1)}bchi_{F_n^{(j)}}(z)dz biggl) =$$

$$= (star)sum_{j=1}^{k-1}overline{g}(c_j)biggl( a lim_{nto +infty}|E_n^{(j)}| + b lim_{nto +infty}|F_n^{(j)}| biggl)(star)$$

where

$$E_n^{(j)}:=biggl{{} bigcup_{ile n-1}biggl(frac{i}{n},frac{i+1/2}{n} biggr) bigcap (c_j,c_{j+1}) biggr{}} mbox{ e } F_n^{(j)}:=biggl{{} bigcup_{ile n-1}biggl(frac{i+1/2}{n},frac{i+1}{n} biggr) bigcap (c_j,c_{j+1}) biggr{}} mbox.$$

These next step is where i think i put to much effort for a proof like this, but i am simply finding the limit of the measure of these two sets.

Lets take $I_n^{(j)}:={ ile n-1:bigl(frac{i}{n},frac{i+1}{n}bigr)subseteq (c_j,c_{j+1}) }$, by definition

$$(star star)E_n^{(j)} supseteq bigcup_{iin I_n^{(j)}}biggl(frac{i}{n},frac{i+1/2}{n}biggr) mbox{ e } F_n^{(j)} supseteq bigcup_{iin I_n^{(j)}}biggl(frac{i+1/2}{n},frac{i+1}{n}biggr) (star star)mbox,$$

and we can observe that for $n ge frac{2}{(c_{j+1}-c_j)}$ we get

$$bigcup_{l=0}^{lfloor n(c_{j+1}-c_j)rfloor-1} biggl( c_j+frac{l}{n},c_j+frac{l+1}{n} biggr) mbox{, with measure } le |(c_j,c_{j+1})|$$

and $exists$ $overline{i}in I_n^{(j)}$ such that

$$frac{overline{i}}{n}in biggl(c_j,c_j + frac{1}{n}biggr)mbox,$$

We can observe that we can obtain our sets

$$ bigcup_{iin I_n^{(j)}}biggl(frac{i}{n},frac{i+1}{n}biggr) mbox,$$

by a translation $le 1/n$ to the right of the sets $bigcup_{l=0}^{lfloor n(c_{j+1}-c_j)rfloor-1} biggl( c_j+frac{l}{n},c_j+frac{l+1}{n} biggr)$. With this process i might loose the last of my intervals, but I loose at most one.

So i can deduce this: $|I_n^{(j)}| ge lfloor n(c_{j+1} – c_j) rfloor -1 ge n(c_{j+1}-c_j)-2$.

Now, using $(star star)$,

$$lim_{nto +infty}|E_n^{(j)}| ge lim_{nto +infty}sum_{iin I_n^{(j)}}biggl| biggl( frac{i}{n},frac{i+1/2}{n} biggr) biggr| = lim_{nto +infty}sum_{iin I_n^{(j)}}frac{1}{2n} = lim_{nto +infty}frac{|I_n^{(j)}|}{2n} ge$$

$$ge lim_{nto +infty}frac{n(c_{j+1}-c_j)-2}{2n} = frac{(c_{j+1}-c_j)}{2}$$

and in the same way

$$lim_{nto +infty}|F_n^{(j)}| ge frac{(c_{j+1}-c_j)}{2} mbox,$$

but since we have also that $|E_n^{(j)} cup F_n^{(n)}|=|E_n^{(j)}|+|F_n^{(j)}|=|(c_j,c_{j+1})|=(c_{j+1}-c_j)$, we can deduce that

$$lim_{nto +infty}|E_n^{(j)}|=lim_{nto +infty}|F_n^{(j)}|=frac{(c_{j+1}-c_j)}{2} mbox.$$

This holds $forall 1ge jge k-1$, so now if we take

$$nge max_{j}biggl{{} frac{2}{(c_{j+1}-c_j)} biggr{}}$$

we can conclude our first series of equalities in $(star)$ with:

$$=lim_{nto +infty}int_{(0,1)}overline{g}u_ndm = sum_{j=1}^{k-1} overline{g}(c_j)frac{(a+b)}{2}(c_{j+1}-c_j)=sum_{j=1}^{k-1}int_{c_j}^{c_{j+1}}overline{g}(z)frac{(a+b)}{2}dz=int_{(0,1)}overline{g}(z)frac{(a+b)}{2}dz mbox,$$

which is exactly what we wanted.