On basic measure and integration

Question

Let $\int_{0}^{1}fg \text{ }d\mathbb{P}=0$, for all $f$ $\in$ $L^{\infty}([0,1],\mathbb{P})$ and $g$ be a fixed function in $L^{1}([0,1],\mathbb{P})$ and where $\mathbb{P}$ is a probability measure in $[0,1]$. Is it true $g=0$ almost everywhere? Attempt: I was trying to use dominated convergence theorem to show $\int|g|^{2}\text{ }d\mathbb{P}=0$, just want to find $f_{n}\rightarrow \overline{g}$ bounded sequence, that I can do using density of $L^{\infty}$ in $L^{1}$, but the problem is $f_{n}$ need not dominated by $L^{1}$ function for applying DCT.

Answer 1

Yes since the dual of $L^1$ is $L^{\infty}$

$T_g(f):=\int_0^1fg$ is a bounded linear functional $T:L^{\infty} \to \Bbb{C}$

and $T_g(f)=0,\forall f \Longrightarrow g=0$

$|T_g(f)| \leq ||f||_{\infty}||g||_1$

Answer 2

Yes. Take $f=I_{g>0}$ to get $\int_{g>0} gdP=0$. Similarly $\int_{g<0} gdP=0$. These give $\int |g|dP=0$ so $g=0$ almost everywhere w.r.t. $P$.