Can every μ-recursive function be defined using a single instance of the μ operator applied to a primitive recursive function?


According to Wikipedia, any μ-recursive function can be expressed as the μ-operator over a primitive recursive function ([source](https://en.wikipedia.org/wiki/%25CE%259C-recursive_function#Normal_form_theorem)):



> 
> A consequence of this result is that
>  any μ-recursive function can be
>  defined using a single instance of the
>  μ operator applied to a (total)
>  primitive recursive function.
> 
> 
> 


From this I conclude that given a μ-recursive function $f(x\_1,\ldots,x\_n)$, I can write it as $\mu y.R(x\_1,\ldots,x\_n,y)$ with $R$ being a primitive recursive function.


If so, let's say there's a function $h(x,y)$ which is μ-recursive but not primitive recursive and its range is $0,1$. As $h$ is μ-recursive, then I can write it as $\mu z.R(x,y,z)$ for some $R$ primitive recursive. However, as $h$'s range is bounded, the 1-bounded μ-recursive operator (which is primitive recursive) over $R$ should give us $h$ as primitive recursive. Absurd! Regarding the existence of $h$, in [this review](http://www.jstor.org/stable/2269168) they show how to construct one of these if I understand correctly.


The question, then, is: what am I misunderstanding in Wikipedia's remark?


Final note: this seems to be an example in Kleene's book Introduction to metamathematics (§58), but I don't see how it is solved.



Representation of μ-recursive functions
The answer is that you have to apply another primitive recursive function after the $\mu$ operator. Specifically, the [Kleene normal form](https://en.wikipedia.org/wiki/Kleene%27s_T_predicate) is that every recursive function $f$ has the form $f(n)=U(\mu x T(e,n,x))$, where both $U$ and $T$ are primitive recursive. The predicate $T(e,n,x)$ asserts that $x$ is the code of a halting computation of program $e$ on input $n$, and the function $U$ extracts the output value from this code. 


It is the step involving $U$ on which your proposed argument flounders.


