From the Riesz representation theorem, we know that the dual of $L^p$ is isomorphic to $L^q$ for $1<p<\infty$, where $q$ is the conjugate index of $q$ (i.e. $1/p+1/q=1$). And if the measure space $(X,\mathcal{X},\mu)$ is $\sigma$-finite, then the dual of $L^1$ is isomrophic to $L^\infty$. But the dual of $L^\infty$ is subtler, The dual $L^\infty$ is isomorphic to the space of signed finitely additive measures on $X$ that are absolutely continuous with respect to $\mu$, which is named as ba space. Why is the dual of $L^\infty$ is not $L^1$? We can simply show that $(L^\infty)^*\neq L^1$, but I want to know why the argument used in $1\leq p<\infty$ doesn't work in $p=\infty$.
Chasing through the proof of the Riesz representation theorem, I notice that the dominated convergence theorem fails in the case $p=\infty$, that is if $(f_n)_{n=1}^\infty$ is a sequence of measurable function dominated by a non-negative function $g\in L^p$ and $(f_n)_{n=1}^\infty$ converges to a limit $f$, then we have $\|f_n\|_{L^p}\to \|f\|_{L^p}$ as $n\to\infty$ if $1\leq p<\infty$. But this is not true if $p=\infty$. Indeed, if we take $f_n=1_{[n,n+1]}$, clearly, we have $f_n\in L^\infty(\mathbb{R}$) and $|f_n|\leq 1$, but $\lim_{n\to\infty}\|f_n\|_{L^\infty}\neq \|f\|_{L^\infty}$ since the LHS is $1$ and the RHS is $0$. Is this the reason?
