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Let $f:[a, b] \to \Bbb R$ be a continuous function such that $f = 0$ almost everywhere; that is, the set $D = \{x \in [a, b]: f(x) \neq 0 \}$ has a measure zero. Prove $f(x) = 0$ for all $x \in [a, b]$.

Here is what I was thinking:

if $f(x_0) \gt 0$, then there exists an interval $I = (x_0 − \delta, x_0 + \delta)$ such that $f(x) \gt 0$ for all $x \in I \cap [a,b]$.

I think that's the idea I want to show, but I'm just having trouble how I can show it.

Robert Shore
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bgj123
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    Your idea is correct: use, that $f=0$ is dense in $[a,b]$, so any neighbourhood of $x_0$ contain its elements. – zkutch Apr 09 '21 at 01:23
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    If $ a<b$ and $x_0\in [a,b]$ and $\delta>0$ then $[a,b]\cap (x_0-\delta,x_0+\delta)$ is an interval of positive length. A basic result about Lebesgue measure is that the measure of an interval is its length. – DanielWainfleet Apr 09 '21 at 02:29

1 Answers1

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You're nearly there. What measure does that set have?

Cameron Williams
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