Show that the set has measure zero [duplicate]

Possible Duplicate:
If $\int_0^x f \ dm$ is zero everywhere then $f$ is zero almost everywhere

If $f$ is integrable on $[a,b]$ and integral from $a$ to $x$ of $f=0$, for every $x \in [a,b]$, show that the set of all $x \in [a,b]$ for which $f(x)$ does not equal $0$ has measure $0$.

Here is what I have so far:

$\displaystyle \int_a^b f(x) \ \mathrm{d}x = 0$ means that the area under the curve of $f(x)$ from $f(a)$ to $f(b)$ is $0$. Hence all the points between $a$ and $b$ lie on the $x$-axis

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marked as duplicate by Byron Schmuland, Jason DeVito, t.b., Nate Eldredge, Martin SleziakAug 7 '12 at 18:22

@mixedmath I don't think the statement that $f$ is non-negative is necessary. The statement as given is required to hold for all $x\in [a,b]$ –  EuYu Aug 6 '12 at 19:49
Define $F(x) = \int_a^x f(t)\, dt.$ for $x\in[a,b]$. Then by your hypothesis, $F(x) = 0$, $x\in[a,b]$. The Lebesgue differentiation on the integral says that $F'(x) = f(x)$ a.e. Your conclusion flows right out of this observation.