I am trying to solve the following question but i am almost positive that i need some result that i don't know. I am free to use any measure theoretic tools. Any help is appreciated.

If $\lambda$ is positive, finite, regular, Borel measure on the real line and $f\in L^1(\lambda )$, suppose that $\int_{(-\infty ,x]}fd\lambda =0, \forall x\in \mathbb{R}$, then show that $f=0$ a.e.

  • $\begingroup$ the derivative of the indefinite integral is f(x) almost every where wrt lambda. $\endgroup$ – Adelafif Jul 30 '15 at 15:19
  • $\begingroup$ Yes i have seen this result somewhere. So the exercise using this theorem is almost trivial. Thanks! $\endgroup$ – user163644 Jul 30 '15 at 15:20
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    $\begingroup$ @Adelafif Well no, the derivative is not $f(x)$. Trivial counterexample: Say $\lambda$ is $2$ times Lebesgue measure. Ok, multiply that by $\chi_{[0,1]}$ to get a finite measure. Then the derivative is $2f(x)$ almost everywhere on $[0,1]$. $\endgroup$ – David C. Ullrich Jul 30 '15 at 16:42
  • $\begingroup$ That theorem would make it trivial, but that theorem does not exist. It still is trivial, using results in that general circle of ideas. Define a measure $\nu$ by $d\nu=f\,d\lambda$. Then there is a theorem that says the norm $||\nu||$ is equal to the total variation of $f$... $\endgroup$ – David C. Ullrich Jul 30 '15 at 16:43
  • $\begingroup$ Rudin, Real and Complex Analysis. Theorem 1.39 (a): Suppose $f: X\to[0, \infty]$ is measurable, $E\in \mathcal{M}$ and $\int_E fd\mu = 0$. Then $f = 0$ a.e. on $E$. If you use this in conjunction with something like Fatou's lemma or Lebesgue's convergence theorem to show that the integral of a limit is zero, I think it should do. The series of functions $f_n = \chi_{(-\infty, x]} f(x)$ $\endgroup$ – Igor Jul 30 '15 at 17:33

Regularity is irrelevant here. Let

$$ G := \{M \in \mathcal{B}\,\mid\, \int_M f \,d\lambda =0\}, $$ where $\mathcal{B}$ is the Borel sigma algebra.

It is easy to see that $G$ is closed under countable disjoint unions.

By letting $x=n\to\infty$ in $0=\int_{(-\infty, x]} f \,d\lambda$, we see $\Bbb{R}\in G$ (dominated convergence), which makes it easy to see that $G$ is close under complements. Hence, $G$ is a $\lambda$ system (also called Dynkin system).

Since $G$ contains the $\pi$ system of all $(-\infty, x]$, which also generates the Borel sigma algebra, we get $G=\mathcal{B}$.

Now let $M=\{x \mid f(x)\geq 0\}$ (and analogously for$\leq$), to get $f=0$ $\lambda$ almost everywhere.


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