Stuck in proving a certain property

Question

I am proving that $\|f\| _{1} =\int_{a}^{b} |f|$ is a norm. but I am stuck in the second part of the first property.

My attempt:

Assume that $\|f\| _{1} = 0,$ then by the definition of $\|f\| _{1}$ and by Proposition 9 on pg.80 (Royden 4th edition), $|f| = 0$ a.e. on $[a,b].$ but then how this leads to that $f = 0.$ I was thinking about using $|f| = f^{+} + f^{-},$ but I do not know how.

Could anyone help me, please?

Answer 1

Assuming that $|f(x_{0})|>0$ for some $x_{0}\in[a,b]$, then there are some $c>0$ and some closed interval $I\subseteq[a,b]$ such that $|f(x)|\geq c$ for $x\in I$, then $\|f\|_{1}\geq\displaystyle\int_{I}|f(x)|dx\geq\int_{I}cdx=c|I|>0$, so with $\|f\|_{1}=0$ it follows that $f(x)=0$ for all $x\in[a,b]$.

Answer 2

For each $x\in[a,b]$, $\bigl\lvert f(x)\bigr\rvert=0\iff f(x)=0$.