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Well, I was trying to prove this, particularly $(2)\Rightarrow (3)$ without using more than the definition of being Riemann integrable (not using density of continuity points or Lebesgue measure 0 for the set of points in which f is not continuous):

Let $f:[a,b]\to\mathbb{R}$ a Riemann integrable function. The following statements are equivalent:

$(1)$ $\int^{b}_{a}|f(x)|dx=0$

$(2)$ If $f$ is continuous at $c \in [a,b]$, then $f(c)=0$

$(3)$ The interior of $X=\{x\in [a,b]: f(x)\neq 0\}$ is empty.

Can someone help me?

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(2) $\implies$ (3)

Assume that $(3)$ is false and $X^o$, the interior of $X$, is non-empty. Then there is a non-empty open interval $I \subset X^o$ such that $|f(c)| > 0$ for every $c \in I$. If $f$ is continuous at any $c \in I$ contradicting condition $(2)$. Otherwise, $f$ must be discontinuous everywhere in $I$ which violates the premise that $f$ is Riemann integrable. Thus if condition $(3)$ is false, so too is condition $(2)$, and $(2) \implies (3).$

(1) $\implies$ (3)

Suppose $X^o$, the interior of $X$, is non-empty. Since $X^o$ is an open set it is a countable union of disjoint non-empty open intervals where $|f(x)| > 0$

$$X^o = \bigcup_{k=1}^\infty (a_k,b_k),$$

and for some finite $n$ we get the contradiction

$$\int_a^b |f(x)| \, dx \geqslant \sum_{k=1}^n \int_{a_k}^{b_k}|f(x)| \,dx > 0.$$

(1) $\implies$ (2)

Suppose $f(c) \neq 0$ and $f$ is continuous at $c$. Then there exists an interval $(c - \delta, c+ \delta) \subset [a,b]$ where $|f(x)| > |f(c)|/2 > 0$ and we get the contradiction

$$\int_a^b |f(x)| \, dx \geqslant \int_{c-\delta}^{c + \delta}|f(x)| \, dx > 2 \delta \frac{|f(c)|}{2} > 0.$$

This should help you get started.

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  • $\begingroup$ I proved all this, the problem is (2)implies(3), I can't find a simple way to prove it. $\endgroup$ – Ángela Flores Mar 7 '17 at 11:42
  • $\begingroup$ I added (2) $\implies$ (3) proved by contradiction. You need to bring in the premise that $f$ is Riemann integrable since there is no information about $\int_a^b |f| = 0$ as a premise for proving this implication. The minimal requirement is that there is no interval where $f$ is everywhere discontinuous. Otherwise you could have (3) false and (2) vacuously true. for functions that are nowhere continuous. $\endgroup$ – RRL Mar 7 '17 at 21:05
  • $\begingroup$ But the OP says not to use $f$ is continuous a.e. $\endgroup$ – zhw. Mar 7 '17 at 21:12

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