We assume that for every real $x$, $L[x]$ only contains countably many reals.


Given a set $X$ of reals, then $L$-ideal generated by $X$ is the smallest set $I$ of reals so that 


1. For any reals $x\in I$ and $y$, $y\in L[x]$ implies $y\in I$; and
2. For any finite $F\subseteq X$, there is a real $z\in I$ so that $F\subseteq L[z]$.


The question is



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> **Question**: Given a null set $X$ only containing $L$-random reals, must the $L$-ideal $I$ generated by $X$ be null?
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Note that, given the set $X$ as in the question, the $L$-upward closure $U\_X=\{y\mid \exists x\in X(x\in L[y])\}$ must be null.



The measure of ideals generated by random reals
The question has a negative answer. The technique is essentially due to Jockusch and Posner.



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> **Proof**: Let $x$ be a real in which every constructible real is recursive. Now $$A=\{r\mid r\mbox{ is Martin-L\" of random relative to }x \wedge x\oplus r\geq\_T x',\mbox{ the Turing jump of }x.\}$$ Then $A$ is null, $r\oplus x$ ranges over the upper cone $\geq\_T x'$, and only contains $L$-random reals. Now for any $z\geq\_T x'$ and $r \in A$, we have that
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> (1). $r\triangle x \in A$; and
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> (2). $(r\triangle x) \oplus r\equiv\_T x\oplus r\geq\_T x'$.
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> So the $L$-ideal generated by $A$ is $\mathbb{R}$.
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