$f\in\R(RR)$ easy

[fu^2]

let $R(RR)={f:RR->RR|int_(-oo)^(+oo)f(x)dx\in\RR}

find a function $f\in\R(RR)$ with $"sup"(f)=+oo

[gugo82]

"fu^2":no, i don't know lebesgue ingration, but i use normal riemann integral

So:

$R(\RR ) :=\{ f:\RR \to \RR :\quad f " is Riemann integrable on the whole real line and " \int_(-oo)^(+oo) f(x) ""x !=\pm oo\}$...

However, my solution works as well with Riemann integral.

Another solution, a more complicated one: choose a number $N\in \NN \setminus \{ 0,1\}$, exactly $N$ distinct points $a_1<\ldots
$f(x):= \{ ((a_1\pi)/x^2, " if " x< a_1), ( 1/\sqrt((x-a_1)(a_2-x)), " if " a_1a_N), (0, " if " x=a_i " for some " 1<=i<=N):} \quad$.

Such an $f$ has $\int_(-oo)^(+oo) f(x)" d"x=(N+1)\pi$, with the integral taken in the generalized Riemann sense.

[fu^2]

no, i don't know lebesgue ingration, but i use normal riemann integral

[gugo82]

What type of integral do you use? Lebesgue or Riemann integral?

Here I suppose you're using Lebesgue integration (if this is the case, you have to specify that every member of $R(\RR)$ is, at least, measurable).
For $x\in \RR$ put:

$f(x):=\{(0, " if " x=0), (1/\sqrt(|x|), " if " 0<|x|<=1), (1/x^2, " if " |x|>=1):}$

Such an $f$ is continuous a.e. (actually in $\RR\setminus \{ 0\}$), hence it's measurable, has $"sup"_\RR f=+oo$ and its Lebesgue integral is two times the improper Riemann integral $\int_0^(+oo) f(x) " d"x$ (whose existence can be proved with a trivial argument), so:

$\int_(-oo)^(+oo) f" d"\ccL =2(\int_0^1 1/\sqrt(x) " d"x+ \int_1^(+oo) 1/x^2 " d"x)=6$.