Let $f:[a,b]\to R$ be an integrable function such that for all $x \in[a,b]$, we have $\int_a^x f(t) \text{dt}=0$. Show that if $f$ is continuous at $x \in [a,b]$, then $f(x)=0$.

My attempt: argue by contradiction.

We assume that $f(x)>0$.

$f$ might not be continuous on the whole interval $[a,b]$, but since it is continuous at point $x\in[a,b]$, then by definition of continuity, $\exists \delta>0 $ such that $\forall t\in(x-\delta,x+\delta)$, $f(t)>0$.

According to what we know, $\int_a^{x+\delta}f(t)\text{dt}=0$, and $\int_a^{x-\delta}f(t)\text{dt}=0 $. Since $f$ is integrable on $[a,x+\delta]$, then it's integrable on the subinterval $[x-\delta,x+\delta]$. This gives us:

$\int_a^{x-\delta}f(t)\text{dt}+\int_{x-\delta}^{x+\delta}f(t)\text{dt}=\int_a^{x+\delta}f(t)\text{dt}$, which implies that $\int_{x-\delta}^{x+\delta}f(t)\text{dt}=0$. However, since $f(t) >0$ for $t\in(x-\delta,x+\delta)$, this cannot be the case. So we have a contradiction.

  • 3
    $\begingroup$ This works, if you add the case where $f(x) < 0$ on some open interval and proceed the same way. Then since $f$ cannot be non-zero on any interval and is continuous, it must be zero. $\endgroup$ – guest196883 Jun 8 '14 at 3:16
  • $\begingroup$ Should be posted as the answer. Anyway, +1 $\endgroup$ – Sandeep Silwal Jun 8 '14 at 3:19
  • $\begingroup$ @SandeepSilwal I'm looking for proof verification, actually :p $\endgroup$ – 3x89g2 Jun 8 '14 at 3:22
  • $\begingroup$ You need to justify, I suppose, why $f>0$ on an interval means that the integral must be positive. If $\inf f=0$, this takes another line to justify. $\endgroup$ – Ted Shifrin Jun 8 '14 at 3:26

HINT: For an easier proof, apply the (first) Fundamental Theorem of Calculus to $F(x)=\displaystyle\int_a^x f(t)dt$.

  • $\begingroup$ OK now I see this is an easier way. Back then we haven't talked about it yet. $\endgroup$ – 3x89g2 Jun 10 '14 at 3:53

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