The problem is to show that for a bounded sequence $\{f_n\}$ in $L^p[a,b]$ that $$ f_n\rightharpoonup f\,\Longleftrightarrow\, \int_E f_n\to\int_E f,\,\, \forall E_\text{(measurable)}\subset [a,b]. $$
Here is my attempt so far:
In class, we were given the example that if $X=L^{p}[a,b]$ with $1\le p<\infty$, then \begin{equation}f_n\rightharpoonup f\,\,\Longleftrightarrow\,\,\forall g\in L^q[a,b]\text{ such that } \int_a^b gf_n\to \int_a^b gf,\end{equation} where $\frac{1}{p}+\frac{1}{q}=1$.
$(\Longrightarrow)$ So let's assume that $f_n\rightharpoonup f$. Observe that for each measurable $E\subset[a,b]$, $\chi_E\in L^{q}[a,b]$ and that the constant function $1\in L^q[a,b]$. Then by our example from class we have $$ \lim_{n\to\infty}\int_E f_n=\lim_{n\to\infty}\int_E 1\cdot f_n =\lim_{n\to\infty}\int_a^b \chi_E\cdot f_n=\int_a^b \chi_E\cdot f=\int_E 1\cdot f=\int_E f. $$
$(\Longleftarrow)$ Now let's assume that $\int_E f_n\to\int_E f$ for all measurable $E\subset[a,b]$. Fix $E=[a,b]$ and let $T$ be a bounded linear operator on $L^p[a,b]$. Since $T$ is bounded, then $T$ is continuous, so we have \begin{equation} \lim_{n\to\infty} T(f_n)=T(\lim_{n\to\infty}f_n) =T(f). \end{equation} But the Riesz Representation Theorem tells us that the bounded linear operators are in one-to-one correspondence with elements of $L^q[a,b]$, so (2) is equivalent to saying $$ \forall\, g\in L^q[a,b],\,\,\int_a^b gf_n=\int_a^b gf. $$ Then by our example we conclude that $f_n\rightharpoonup f$.
The only problem I see with the reverse direction is that I need $\lim_{n\to\infty} f_n=f$ pointwise, so that when I bring the limit inside the continuous operator I can conclude that $T(\lim_{n\to\infty}f_n) =T(f)$. Is this fixable?
UPDATE: Here is the new and improved reverse argument!
$(\Longleftarrow)$ Now let's assume that $\int_E f_n\to\int_E f$ for all measurable $E\subset[a,b]$. Fix $E=[a,b]$ observe that $f_n\to f$, except on a set of measure zero, say $F$. Let $T$ be a bounded linear operator on $L^p[a,b]$. Since $T$ is bounded, then $T$ is continuous, so on $E\setminus F$ we have \begin{equation} \lim_{n\to\infty} T(f_n)=T(\lim_{n\to\infty}f_n) =T(f).\qquad(2) \end{equation} But the Riesz Representation Theorem tells us that the bounded linear operators are in one-to-one correspondence with elements of $L^q[a,b]$, so (2) is equivalent to saying \begin{equation} \forall\, g\in L^q[a,b],\,\,\lim_{n\to\infty}\int_{E\setminus F} gf_n=\int_{E\setminus F} gf.\qquad(3) \end{equation} Since $F$ is of measure zero we have $$ \forall\, g\in L^q[a,b],\,\,\lim_{n\to\infty}\int_a^b gf_n=\lim_{n\to\infty}\int_{E\setminus F} gf_n\overset{(3)}{=}\int_{E\setminus F} gf=\int_a^b gf. $$ Then by our example in class we conclude that $f_n\rightharpoonup f$.
Unfortunately, a sequence can be weakly convergent without being pointwise convergent $(f_n(x))=(\sin(n\pi x))$ on the unit interval (which is not convergent except at $0$).
However we can show that for each simple function $g$ (linear combination of characteristic functions of measurable sets), $$\lim_{n\to +\infty}\int_{(a,b)} f_ngdx= \int_{(a,b)} fgdx.$$ Using boundedness of the sequence $(f_n)_n$ and Hölder's inequality, we can extend this of each $g\in \mathbb L^q$.