Famously, Solovay showed that, if $\textrm{ZFC}$ plus $\textrm{IC}$ (the existence of an inaccessible cardinal) is consistent, then so is $\textrm{ZF}$ plus $\textrm{DC}$ (dependent choice) plus $\textrm{LM}$ (all subsets of $\mathbb{R}$ are Lebesgue measurable). And Shelah showed that, conversely, the consistency of $\textrm{ZF}+\textrm{DC}+\textrm{LM}$ implies the consistency of $\textrm{ZFC}+\textrm{IC}$.


Since $\textrm{LM}$ is a statement of third-order arithmetic, it makes sense to consider it in theories weaker than $\textrm{ZF}$ set theory. I am curious what is known about the consistency strength of $\textrm{DC} + \textrm{LM}$ in such weaker contexts. Of specific interest to me is the consistency strength of $\textrm{DC}+\textrm{LM}$ over the base logic of a boolean topos (= higher-order classical logic). 



How strong is "all sets are Lebesgue Measurable" in weaker contexts than ZF?
The consistency of ZFC + IC is perhaps a little bit too much to ask, but I believe the next best thing is true:


**Conjecture.** Every boolean topos1 with dependent choice in which every set of reals is Lebesgue measurable contains a well-founded model of ZFC. In fact, every real is contained in a well-founded model of ZFC.


The proof that Con(ZF + DC + LM) implies Con(ZFC + IC) shows that in any model $V$ of ZF + DC + LM, $\aleph\_1$ (of $V$) is an inaccessible cardinal in the constructible universe $L$ and therefore the inner model $L$ satisfies ZFC + IC. In fact, it shows that $\aleph\_1$ is inaccessible in $L[a]$ for every real $a$ and therefore $L\_{\aleph\_1}[a]$ is a well-founded model of ZFC that contains $a$. I will now argue that we can still make sense of "$L\_{\aleph\_1}[a]$ is a well-founded model of ZFC that contains $a$" in a boolean topos with dependent choice and I will explain why I believe why this is true in a boolean topos with dependent choice in which all sets of reals are Lebesgue measurable.


Since there are no actual material sets around, we must first find a substitute. A boolean topos can still make sense of HC, the collection of all hereditarily countable sets, by amalgamating all countable well-founded extensional structures $(\mathbb{N},E)$. More precisely, one can show that any two such structures have at most one transitive embedding between them and that any two of them can be amalgamated into a third. (All that is needed for this is arithmetic transfinite recursion.) Then the limit of this directed system of countable well-founded extensional structures under transitive embeddings is the required HC.


Once we have HC, we can do a bit of set theory in there, thinking of its elements as actual material sets. In fact, assuming dependent choice, HC is a very nice model: it satisfies comprehension, replacement, choice and the basic combinatorial axioms but it doesn't have powersets, of course, since every set is countable. The ordinals of HC form a well-ordering that we will call $\aleph\_1$. Working in HC, one can construct $L\_\eta[a]$ for every $\eta \in \aleph\_1$ and every $a \subseteq \omega$. Therefore, one can make sense of the substructure $L\_{\aleph\_1}[a]$ of HC, which is a well-founded model of a fragment of ZFC + $V = L[a]$ and the question whether $L\_{\aleph\_1}[a]$ is a model of ZFC makes sense.


The key ingredient that connects these ideas with Lebesgue measurability is the following theorem of Raisonnier:


**Theorem.** Assume ZF + DC. If there is an uncountable well-orderable subset of $\mathbb{R}$ then there is a non-measurable set of reals.


I don't know whether this goes through in a boolean topos with dependent choice. However, since this is a theorem of "ordinary mathematics," I conjecture that it does! The theorem does use some "fancy objects" such as rapid ultrafilters but these do make sense in a boolean topos and, in the presence of dependent choice, so does Lebesgue measure. I could be wrong, but I can't see any immediate obstructions to Raisonnier's Theorem in a boolean topos with dependent choice.


Now, the $a$-constructible reals $\mathbb{R}^{L[a]} = \mathbb{R}\cap L\_{\aleph\_1}[a]$ form a well-orderable set of reals since $L\_{\aleph\_1}[a]$ has a definable wellordering. Therefore, assuming that Raisonnier's Theorem goes through, in any boolean topos with dependent choice where all sets of reals are Lebesgue measurable, it must be the case that $\mathbb{R}^{L[a]}$ is countable for every $a \subseteq \omega$. Then, the usual argument that shows that $\aleph\_1$ is inaccessible in $L[a]$ goes through to show that $L\_{\aleph\_1}[a]$ is a model of ZFC.


To summarize, at Alex's request, the above sketches a proof of the following:


**Theorem.** In a boolean topos with dependent choice where all well-orderable sets of reals are countable, every real is contained in a well-founded model of ZFC.


So the truth of the initial conjecture rests only on the truth of Raisonnier's Theorem in a boolean topos with dependent choice. Note that the conclusion is rather strong. Since the models are well-founded, all $\Sigma^1\_2$ statements that are true in such models are also true in the ambient topos. In particular, every such topos proves that ZFC is not only consistent but also $\Sigma^1\_2$-sound.


To get the relative consistency of ZFC+IC as in Shelah's Theorem, we would need to continue the construction of $L$ past $\aleph\_1$ and show that this leads to a model of ZFC in the limit. It is unlikely that this is possible without some kind of additional completeness assumptions on the topos. 




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1For the sake of brevity, by "topos" I always mean "topos with a natural number object."


