Norm of Linear functional on $C[0,1]$

Question

Let $\mu$ be a finite measure on $[0,1]$. Let $T(f)=\int fd\mu$ is a bounded linear functional on $C[0,1]$, then $\left\Vert T\right\Vert =\mu([0,1])$. I know the proof of one direction: $$\left|T(f)\right|=\left|\int fd\mu\right|\le\int\left\Vert f\right\Vert _{\infty}d\mu=\left\Vert f\right\Vert _{\infty}\mu([0,1])$$ so $\left\Vert T\right\Vert \le\mu([0,1]).$ But how to show the other direction?

Answer 1

You take $f=1$. Then $\lVert f\rVert_\infty=1$ and $\bigl\lvert T(f)\bigr\rvert=\mu\bigl([0,1]\bigr)$.

Answer 2

Hint:

You need to find a function $g\in C[0,1]$ such that $\|g\|_\infty=1$ and $|T(g)|=\mu([0,1])$.

This would then imply $\|T\|=\mu([0,1])$.