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I wanna prove that

"if $f: \mathbb{R}^n \to \mathbb{R}$ is continuous and satisfies $f=0$ almost everywhere (in the sense of Lebesgue measure), then, $f=0$ everywhere."

I am confident that the statement is true, but stuck with the proof. Also, is the statement true if the domain $\mathbb{R}^n$ is restricted to $\Omega \subseteq \mathbb{R}^n$ that contains a neighborhood of the origin "$0$"?

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    @sos440: Sorry, I exaggerated. :)2012-11-06

4 Answers 4

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Here is a generalization of the result that you want:

Theorem: Let $f,g$ be two continuous functions such that $f = g$ a.e. Then $f = g$ everywhere.

Proof: Let $E$ be the set of all $x$ such that $f(x) \neq g(x)$. Suppose $E$ is not empty and so contains some $x$. Then $E$ being the complement of a closed set is open and so we can find $\epsilon > 0$ such that $B_\epsilon(x) \subseteq E$. But now this means $0 < \mu(B_\epsilon(x)) \leq \mu(E)$ contradicting $\mu(E) = 0$. It follows that $E$ has to be empty so that $f = g$ everywhere.

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    @Yang The fact that $\{x\mid f(x) = g(x)\}$ is closed holds for mappings on Hausdorff spaces.2016-03-27
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A set of measure zero has dense complement. So if a continuous function zero on a set of full measure, it is identically zero.

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Since $f$ is continuous, if $f(\hat{x}) \neq 0$, then there exists a $\delta>0$ such that $|f(x)|> \frac{1}{2}|f(\hat{x})|$ for $x \in B_\infty(\hat{x},\delta)$. Since $m(B_\infty(\hat{x},\delta)) = (2 \delta)^n>0$, we see that if $f(\hat{x}) \neq 0$, there exists a set of positive measure on which $f$ is non-zero.

Hence if $f$ is zero a.e., it must be zero everywhere.

(I choose the '$\infty$' ball so I could compute the measure easily.)

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    The function $x \mapsto |f(x)|$ is continuous at $\hat{x}$ so if I let $\epsilon = {1 \over 2} |f(\hat{x})|$ (which is >0 by assumption), then there is some \delta>0 such that if $x \in B_\infty(\hat{x},\delta)$ then |f(x)-f(\hat{x})| < \epsilon. Since $|a-b| \ge |a|-|b|$, we have |f(\hat{x})| < \epsilon + |f({x})|, or |f(x)| > |f(\hat{x})| - \epsilon. Now substitute the chosen value of $\epsilon$.2015-10-26
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This is another simpler but longer approach.

Notice that it is enough to show that $|f|=0$ everywhere, so let's assume that $f$ is nonnegative.

Since $f=0$ a.e. it follows that $\int_{\Bbb R^n} f=0.$ Divide the entire space $\Bbb R^n$ in nonoverlapping cubes of side length 1, say $\{I_k\}_{k\in\Bbb N}$, then $0=\int_{\Bbb R^n} f=\sum_{k\in\Bbb N} \int_{I_k} f$ and then $\int_{I_k} f=0$ for each $k\in\Bbb N$. That's the key of this proof.


Lemma. Let $f:\Bbb R^n\to\Bbb R$ be a continuous nonnegative function. Let $I=[a^1,b^1]\times\cdots\times [a^n,b^n]$ be an interval. If $\int_I f=0,$ then $f(x)=0$ for each $x\in I$.

Proof. The proof is by induction on $n$.

If $n=1$ it is just this.

Suppose that the result holds for $1,\ldots,n-1$. Notice that Fubini's theorem is applicable, so $\newcommand{\d}{\mathrm{d}} \newcommand{\x}{\mathbf{x}} \int_I f(\x)\d\x=\int_{a^n}^{b^n}\left[\int_{a^{n-1}}^{b^{n-1}}\cdots \int_{a^{1}}^{b^{1}} f\left(x^1,\ldots,x^n\right)\d x^{1}\cdots \d x^{n-1}\right]\d x^{n}.\tag{1}\label{eqi}$ Define $K:\left[a^n,b^n\right]\to\Bbb R$ by $K(t)=\left[\int_{a^{n-1}}^{b^{n-1}}\cdots \int_{a^{1}}^{b^{1}} f\left(x^1,\ldots,t\right)\d x^{1}\cdots \d x^{n-1}\right].$ The LHS of \ref{eqi} is $0$, so $K$ is $0$ a.e. in $\left[a^n,b^n\right]$. Since $K$ is continuous in $\left[a^n,b^n\right]$, by our hypothesis follows that $K$ is $0$ identically in $\left[a^n,b^n\right]$.

Now, fix $t\in \left[a^n,b^n\right]$. By Fubini again $\int_{[a^1,b^1]\times\cdots\times [a^{n-1},b^{n-1}]} f\left(x^1,\ldots,x^{n-1},t\right)\d \left(x^1,\ldots,x^{n-1}\right)=K(t)=0,$ by our induction hypothesis, it follows that $f\left(x^1,\ldots,x^{n-1},t\right)=0$ for each $\left(x^1,\ldots,x^{n-1}\right)\in [a^1,b^1]\times\cdots\times [a^{n-1},b^{n-1}]$. Since $t\in \left[a^n,b^n\right]$ is arbitrary it follows that $f\left(x^1,\ldots,x^{n}\right)=0$ for each $\left(x^1,\ldots,x^{n}\right)\in I$, as we wanted.


Then using the Lemma, it follows that $f$ is $0$ everywhere in each $I_k$ and therefore $f$ is identically $0$.

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    Nice answer! I would rather use subscripts than superscripts though!2013-07-27