I know that the tricky part is to prove that the map $\alpha$ that takes any $f \in L^{\infty}[a,b]$ ans sends it to $K_f$, where $K_f(g) = \int_a^b f(x)g(x)dx$ and $g \in L^1[a,b]$ is surjective and isometric. The rest I have already done it.
I have seen a couple of proofs on this site about this fact using the famous Radon–Nikodym theorem. However, they usually work with abstract spaces $X$ and measures $\mu$. In my case it suffices to work with $X=[a,b]$ and with the Lebesgue measure. Thanks in advance.
$X$ has finite mesure so $L^2 \subset L^1$ and thus every linear form $\ell$ on $L^1$ is (by Hilbert space magic) given on $L^2$ by the integration against some $L^2$ function $f$.
Assume that $|\ell(g)| \leq M$ if $\|g\|_{L^1} \leq 1$. Let $S=\{f > M+1\}$. Then $|\ell(1_S)| \leq M\lambda(S)$, but $\int{f1_S} \geq \lambda(S) (M+1)$. Thus $\lambda(S)=0$ and it follows by a similar argument that $|f| \leq M+1$ almost everywhere, thus $f$ is in $L^{\infty}$. Now $\ell$ and the integral against $f$ coincide on $L^2$ (a dense subset of $L^1$) and they’re both continuous so they’re equal.
So $f \longmapsto K_f$ is a bijection. The above argument (suitably modified) shows that $\|f\|_{\infty} \leq \|K_f\|_{(L^1)^*}$. But it’s easy to see that $\|K_f\|_{(L^1)^*} \leq \|f\|_{\infty}$.