Under what condition does $\int_a^b|f(x)|=0$ imply $f=0$? ( at least a.e.). I'm working through Real Analysis by Royden & Fitzpatrick, and on the first section on $L^p$ spaces they always skip the last property of norms without much commentary. Namely, that $||f||=0 \iff f=0$.
For example, they discuss a norm of $L^1$ given by $||f||=\int_a^bx^2|f(x)|dx$. To show $||f||=0 \implies f=0$, they just claim $\int_a^bx^2|f(x)|dx=0$ gives $f=0$ a.e. 
Unfortunately, I do not see how this is immediately obvious, neither from the definition of Riemann or Lebesgue integration. Is this somehow trivial for non negative functions? I've seen some proofs of this for continuous functions, but $f\in L^1$ and is not necessarily continuous here.
 A: (1) if $f$ is measurable, and $\int_a^b |f(t)|\;dt = 0$, then $f(t) = 0$ for almost all $t$.
Proof.  Suppose $|\{t : f(t) \ne 0\}| > 0$.  (I used $|\cdot|$ for Lebesgue measure.)  Then $|\{t : |f(t)|>0\}| > 0.$  But
$$
\{t : |f(t)|>0\} = \bigcup_{n \in \mathbb N} \left\{t : |f(t)| > \frac{1}{n}\right\}
$$
so by countable additivity, there exists $n$ such that
$\left|\left\{t : |f(t)| > \frac{1}{n}\right\}\right| > 0$.  Thus
$$
\int_a^b |f(t)|\;dt \ge \left|\left\{t : |f(t)| > \frac{1}{n}\right\}\right|\cdot\frac{1}{n} > 0.
$$
(2) If $f$ is measurable, and $\int_a^b t^2|f(t)|\;dt = 0$, then $f(t) = 0$ for almost all $t$.
Applying (1) to the function $t^2|f(t)|$, we get $t^2|f(t)|=0$ for almost all $t$.  The points where $f(t)$ are zero, together with (possibly) one point where $t^2=0$, give us the points where $t^2 |f(t)| = 0$.  So from $t^2|f(t)|=0$ a.e. we get $f(t)=0$ a.e.
A: Let $a\leq t\leq b$.
As $x^2|f(x)|\geq0$,
$$0\leq\int_a^t x^2|f(x)|dx\leq\int_a^b x^2|f(x)|dx$$
So, for any $a\leq t\leq b$,
$$\int_a^t x^2|f(x)|dx=0$$
Now differentiate it and you get
$$t^2|f(t)|=0\quad (a\leq t\leq b)$$
$$|f(t)|=0\quad (a\leq t\leq b, t\ne0)$$
$$\therefore f(t)=0\quad (a\leq t\leq b, t\ne0)$$
