Suppose $f$ is R.integrable on a closed interval $[a,b]$ and that $\displaystyle\int_a^x f=0$ for all values of $x$ within this closed interval. Then how can I start to show that this means that $f$ vanishes almost everywhere?

I need a small push in the correct direction to get me started please.

  • $\begingroup$ Show that the sets on which $f$ vanish form a $\sigma$-algebra, and that this $\sigma$-algebra contains all rational intervalls in $[a,b]$. $\endgroup$ – Oles Wohnzimmer Aug 9 '16 at 13:25
  • $\begingroup$ Have you learned Dykins $\pi$-$\lambda$ theorem? $\endgroup$ – Xiao Aug 9 '16 at 14:21
  • $\begingroup$ I am not familiar with that theorem $\endgroup$ – Marshant Aug 9 '16 at 15:07

We must have $\int_{x_1}^{x_2}f(t)dt = \int_{a}^{x_2} f(t)dt - \int_{a}^{x_1}f(t)dt = 0 $ for every $x_1,x_2$ with $x_1 < x_2$ lying in $[a,b].$

Since $f$ is Riemann integrable it is continuous at almost every point of $[a,b]$.

So it is sufficient to show that $f$ is $0$ at every point of continuity in $[a,b].$

If there is an $x_0 \in [a,b]$ such that $f$ is continuous at $x_0$ and $f(x_0) > 0$ then we can find $x_1,x_2$ with $x_1 < x_2$ with $x_0 \in [x_1,x_2] \subset [a,b]$ such that $f(x) > f(x_0)/2 > 0$ on $[x_1,x_2]$. This implies $\int_{x_1}^{x_2} f(t) dt > 0$, a contradiction. There cannot be an $x_0$ at which $f$ is continuous and $f(x_0) < 0$ for similar reasons.


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