Existence of Real-Valued Functions Satisfying Certain Properties.

(1)Is there a real-valued function f satisfies

the set {(x,f(x)),x belongs to R} is a second category subset of R2?

(2)Is there a real-valued function f satisfies

the set {(x,f(x)),x belongs to R} is non-measurable in the Lebesgue sense?

Re: a question about real-valued functions on R

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Originally Posted by

**mathabc** Is there a real-valued function f satisfies the set {(x,f(x)),x belongs to R} is non-measurable in the Lebesgue sense?

__Hint__: For any $\displaystyle A\subset \mathbb{R}$ non Lebesgue measurable, choose the indicator function $\displaystyle \chi_A:\mathbb{R}\to \mathbb{R}$

$\displaystyle \chi_A (x)=\begin{Bmatrix} 1 & \mbox{ if }& x\in A\\0 & \mbox{if}& x\not \in A\end{matrix}$

Quote:

Is there a real-valued function f satisfies the set {(x,f(x)),x belongs to R} is a second category subset of R2?

__Hint__: Use the Baire's Category Theorem: *Every complete metric space is of second category*.

Re: a question about real-valued functions on R

... and $\displaystyle \Gamma (I)=\Delta=\{(x,x):x\in\mathbb{R}\}$ is complete.

And with this, **we** end the lively conversation. :)

Re: a question about real-valued functions on R

Quote:

Originally Posted by

**FernandoRevilla** __Hint__: For any $\displaystyle A\subset \mathbb{R}$ non Lebesgue measurable, choose the indicator function $\displaystyle \chi_A:\mathbb{R}\to \mathbb{R}$

$\displaystyle \chi_A (x)=\begin{Bmatrix} 1 & \mbox{ if }& x\in A\\0 & \mbox{if}& x\not \in A\end{matrix}$

Just an observation: The graph of that function is a subset of a set of measure 0 (the line y=1).

Re: a question about real-valued functions on R

Quote:

Originally Posted by

**Jose27** Just an observation: The graph of that function is a subset of a set of measure 0 (the line y=1).

Right, which implies $\displaystyle \mu^* (\Gamma (1_A))=0$ if you consider the complection $\displaystyle \mu^*$ of $\displaystyle \mu$ .

P.S. I suppose you meant a subset of the union of the lines $\displaystyle y=0$ and $\displaystyle y=1$ .