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Suppose that $a \lt b$ and that $f:[a,b] \to \mathbb{R}$ is bounded. Then prove that:

  • if $f$ is continuous at $x_0 \in [a,b]$ and $f(x_0) \neq 0$ then $$(L) = \int_{a}^{b} \vert f(x) \vert dx \gt 0\text{; and}$$

  • if $f$ is continuous on $[a,b]$ then $$\int_a^b \vert f(x) \vert dx = 0$$ if and only if $f(x) = 0$ for all $x \in [a,b]$.

Toby Bartels
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K. Gibson
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    The second item follows pretty easily from the first. By the way, the hypothesis that $f$ is bounded is not needed. On the other hand, the first result needs something to guarantee that the integral exists. If the integral is a Lebesgue integral, allowed to be infinite, then we need only that $f$ is measurable. We could also take the integral to be a lower Darboux integral, which always exists if allowed to be infinite, and is always $\leq$ any integral that exists, so that's a fairly strong result. – Toby Bartels Nov 07 '19 at 18:20

1 Answers1

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HINT

Since if is continuous at $x_0$ then $|f|$ is contintinuous at $x_0$ so for $\epsilon=\frac{|f(x_0)|}{2}$ exists $\delta>0$ such that $|f(x)| >\frac{|f(x_0)|}{2}>0,\forall x \in (x_0-\delta,x_0+\delta)$

So $$\int_a^b|f(x)|dx \geq \int_{x_0-\delta}^{x_0+\delta}|f(x)|dx >\frac{|f(x_0)|}{2}2\delta>0$$

Marios Gretsas
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  • its struggle to continue – K. Gibson Nov 07 '19 at 18:19
  • @K.Gibson i edited – Marios Gretsas Nov 07 '19 at 18:22
  • @K.Gibson : Let's put Marios's result like this: $|f(x)|\geq g(x)$, where $g(x)=|f(x_0)|/2$ for $x_0-\delta<x<x_0+\delta$, and $g(x)=0$ otherwise. Note that $g$ is piecewise constant, so you should know how to integrate it. And you should know that integrals respect inequalities. So what does that tell you about the integral of $|f|$? – Toby Bartels Nov 07 '19 at 18:24