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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

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    Perhaps we are close to the point where nate or someone could post these conclusions as an answer?2011-10-21

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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 $g(x) = 0$.

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    Looks OK to me.2011-10-21