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Let $D\subset\left[ 0,1\right] $ be a dense set, and $\mu$ Lebesgue measure on $\left[ 0,1\right] .$

Suppose $f:\left[ 0,1\right] \rightarrow\left[ 0,1\right] $ is continuous at each point in $D.$ Let $\overline{G}$ be the closure of the graph of $f$ on $\left[ 0,1\right] ^{2}.$

Is it true that $\mu\left\{ x_{0}:\overline{G}\cap\left\{ x=x_{0}\right\} \text{ is infinite}\right\} =0?$

Robert
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2 Answers2

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Much worse than this is possible, which you can see by using #1 & #2 below.

1. There exist functions $f:{\mathbb R} \rightarrow {\mathbb R}$ such that the graph of $f$ is dense in ${\mathbb R}^2.$ (In fact, this can even be the case for a Baire $2$ function.)

See, for example, these StackExchange questions:

Graph of discontinuous linear function is dense

Does there exist a function $f:[0,1] \to[0,1]$ such its graph is dense in $[0,1]\times[0,1]$?

graph is dense in $\mathbb{R}^2$

2. Given any function $f:{\mathbb R} \rightarrow {\mathbb R}$, there exists a subset $D$ of $\mathbb R$ such that $D$ is dense in $\mathbb R$ and $f$ restricted to $D$ is continuous.

For this result, see the following:

Every real function has a dense set on which its restriction is continuous [mathoverflow question]

Jack B. Brown, A measure theoretic variant of Blumberg's theorem, Proceedings of the American Mathematical Society 66 #2 (October 1977), 266-268.

google search for "Blumberg" and "dense" and "continuous"

Community
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Dave L. Renfro
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  • I think the OP is only considering functions which have a dense (and hence comeager) set of points of continuity. Blumberg's theorem only gives the continuity of any function restricted to some dense set. – hot_queen Jan 10 '14 at 22:43
  • Wow, I didn't even notice that this could have been the interpretation, but in thinking about it now I tend to agree with you. Maybe the OP can reword it as "continuous at each point in $D$" if this is the intended meaning. – Dave L. Renfro Jan 10 '14 at 23:06
  • Thanks Dave, this is interesting and I was not aware of this result; hot_queen is right I meant continuous at every point in D. – Robert Jan 11 '14 at 19:06
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Here's a counterexample: Let $D$ be the complement of a fat Cantor set $C \subseteq [0, 1]$ so $D$ is dense. Construct $f:[0, 1] \rightarrow [0, 1]$ such that $f$ is zero on $D$ and the graph of $f$ is dense in $C \times [0, 1]$.

hot_queen
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  • I dont quite get the question. what is ${x_0: \bar{G}\cap {x=x_0} \text{is infinite}}$? – Lost1 Jan 10 '14 at 20:27
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    I guess it is the set of those $x \in [0, 1]$ such that the vertical section through $x$ meets $\overline{G}$ at infinitely many points. – hot_queen Jan 10 '14 at 20:29
  • I see. I didn't know about these functions that have dense graph, thanks! – Robert Jan 11 '14 at 19:09