Does $f(x)$ is continuous and $f = 0$ a.e. imply $f=0$ everywhere?

Question

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 "$0$"?

Answer 1

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.

<p><strong>Proof:</strong> 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 $\epsilon &gt; 0$ such that $B_\epsilon(x) \subseteq E$. But now this means
$$0 &lt; \mu(B_\epsilon(x)) \leq \mu(E)$$
contradicting $\mu(E) = 0$. It follows that $E$ has to be empty so that $f = g$ everywhere.</p>

Answer 2

A set of measure zero has dense complement. So if a continuous function zero on a set of full measure, it is identically zero.

Answer 3

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.)

Answer 4

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.

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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.

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

Answer 5

Let $f:\mathbb{R}^n\rightarrow\mathbb{R}$ be a continuous function. If $f=0$ a.e. then $f=0$.
Proof. Let $f$ be as stated. Then there is a measurable set $N$ such that $f(x)=0$ for all $x\in\mathbb{R}^n-N$ and $\mu(N)=0.$
Let $x_0\in N.$ Since $\mu(N)=0$, there is no open ball contained in $N$, otherwise, let's say $B\subseteq N$, one has $0<\mu(B)\leq\mu(N)=0$. Hence for all $\varepsilon>0$, $B(x_0,\varepsilon)\not\subseteq N$. This implies that for all $n\in\mathbb{N}$ there exists $x_n\in N^c$ such that $x_n\in B(x_0,1/n)$, and thus $(x_n)\rightarrow x_0$. Moreover, since $x_n\in N^c,$ $f(x_n)=0$ for all $n\in\mathbb{N}.$
Hence by the continuity of $f$, $f(x_0)=f(\lim_{n\to\infty} x_n)=\lim_{n\to\infty} f(x_n)=0$.