Let $(\Omega, \mathcal F, \mu)$ be a $\sigma$-finite measure space. We define a map $$ T : L^1(\Omega, \mu, \mathbb R) \to (L^\infty(\Omega, \mu, \mathbb R))^* $$ by $$ (T u) (f) := \int_{\Omega} uf \ \mathrm d \mu \quad \forall u \in L^1(\Omega, \mu, \mathbb R), \forall f \in L^\infty(\Omega, \mu, \mathbb R). $$
Is $T$ injective?
My attempt Let $u \in L^1(\Omega, \mu, \mathbb R)$ such that $$ \int_{\Omega} uf \ \mathrm d \mu = 0 \quad \forall f \in L^\infty(\Omega, \mu, \mathbb R). $$
Let $\Omega_n := \{|u| \le n\}$ and $f_n := u 1_{\Omega_n}$. Then $f_n \in L^\infty(\Omega, \mu, \mathbb R)$ and $f_n \to u$ pointwise everywhere. Then $$ \int_{\Omega} uf_n \ \mathrm d \mu = 0 \quad \forall n. $$
It's unfortunate that we can not apply DCT.
As @ArcticChar remarked in a comment, $T$ is the canonical injection from $L^1(\Omega, \mu, \mathbb R)$ to its bidual, so it is in fact an isometric isomorphism. This claim can be found at page 8 of Brezis's Functional Analysis.