real analysis - Find Borel functions $f,g$ that agree on a dense subset of R but not at $lambda$-almost every $xin R$

Here is the homework question, verbatim:

Find Borel functions $f,g: \mathbb{R} \to \mathbb{R}$ that agree on a dense subset of $\mathbb{R}$ but are such that $f(x) \neq g(x)$ holds at $\lambda$-almost every $x\in \mathbb{R}$

I interpreted the latter part to mean, "... holds $\lambda$-almost everywhere in $\mathbb{R}$."

I also understand $\lambda$ to actually be $\lambda^{\ast}$ - Lebesgue outer measure.

I also think the question says that $f(x) \neq g(x)$ holds $\lambda$-almost everywhere in $\mathbb{R}$ and means that for only a set of outer Lebesgue measure zero is it true that $f(x) = g(x)$. This latter set happens to be the dense subset of $\mathbb{R}$ mentioned in the problem.

So what I have so far is that $f(x) = g(x)$ on a set $A$ s.t. $\lambda \big(A:=\{x \in \mathbb{R} : f(x) = g(x) \}\big)=0$. So $A$ is dense, meaning that for any $x \in \mathbb{R}$, any neighborhood $N(x,\;\;\;) \ni $ (at least one point from $A$). This to me means that, since the interior of $int \; (A^c) = \varnothing$, only the points ${}^{\pm}\infty$ of the extended real number line is where these functions agree. But I don't see how $\{{}^-\infty\}$ and $\{{}^+\infty\}$ can be dense....?

So does this mean two different functions (classes of functions?) that only share one or both infinite limits?

Thanks much for any guidance!

nate

Answer

Thanks all for the help with this - it may yet need editing? @Gerry ?

So set $g(x)=0$ identically and treat $f(x)$ as an indicator/characteristic function on the set where $f(x) \neq g(x)$. The dense subset of $\mathbb{R}$ where the two functions are equal has the properties, $$f(x) = g(x) \longleftrightarrow \lambda^{\ast}(\left\{ x \in \mathbb{R} : f(x) = g(x) \right\} ) = \lambda^{\ast}\big(H^c:= dense\;\;subset\;\;of\;\;\mathbb{R} \big) = 0,$$ and the function is defined as, $$f(x) = \left\{ \begin{array}{ll} 1,& x \in H \subseteq \mathbb{R} \\ 0, & x \notin H \end{array} \right. .$$ So choose $H$ s.t. $f(x) \neq 0\;\;a.e.$.

Let $H:=\mathbb{R}\backslash \mathbb{Q} \Longleftrightarrow H^c = \mathbb{Q}$, where $\mathbb{Q}$ is dense (and countable). Then the indicator function becomes, $$f(x) = \left\{ \begin{array}{ll} 1, & x \in \mathbb{R}\backslash \mathbb{Q} \\ 0, & x \in \mathbb{Q} \end{array} \right. .$$ As a check then, on $\mathbb{Q}$, $f(x) = 0$ and $g(x) = 0$ (though $g(x) = 0$, being identically 0 implies it is the additive identity of the underlying group, and $f(x)=0$ implies that $\left[f(x)\right] = 0$ as a class of functions - are they really the same?).

On $\mathbb{R} \backslash \mathbb{Q}$, $f(x) = 1$ and $g(x) = 0$.