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This is a follow up question to MO question "Idempotent measures on the free binary system?". Let $(A,*)$ be the free binary operation on one generator which satisfies the left self distributive law: $a * (b * c) = (a * b)* (a* c)$. Is there a finitely additive probability measure $\mu$ on $A$ such that $\mu * \mu = \mu$?

Here $\mu * \nu$ is defined by $$ \mu * \nu (X) = \int \int \mathbf{1}_{*^{-1}(X)} (x,y)\ d\mu (x) \ d \nu (y). $$ Note: because of the asymmetry of the left self distributive law, the above order of integration is important (it is possible to show that there is an idempotent with the opposite order of integration). I conjecture a positive answer (this would follow from a positive answer for the free binary system on one generator) and this question may be easier to resolve than the other.

This is a follow up question to MO question "Idempotent measures on the free binary system?". Let $(A,*)$ be the free binary operation on one generator which satisfies the left self distributive law: $a * (b * c) = (a * b)* (a* c)$. Is there a finitely additive probability measure $\mu$ on $A$ such that $\mu * \mu = \mu$?

Here $\mu * \nu$ is defined by $$ \mu * \nu (X) = \int \int \mathbf{1}_{*^{-1}(X)} (x,y)\ d\mu (x) \ d \nu (y). $$ Note: because of the asymmetry of the left self distributive law, the above order of integration is important (it is possible to show that there is an idempotent with the opposite order of integration). I conjecture a positive answer (this would follow from a positive answer for the free binary system on one generator) and this question may be easier to resolve than the other.

This is follow up question to MO question "Idempotent measures on the free binary system?" Let $(A,*)$ be the free binary operation on one generator which satisfies the left self distributive law: $a * (b * c) = (a * b)* (a* c)$. Is there a finitely additive probability $\mu$ on $A$ such that $\mu * \mu = \mu$?

Here $\mu * \nu$ is defined by $$ \mu * \nu (X) = \int \int \mathbf{1}_{*^{-1}(X)} (x,y)\ d\mu (x) \ d \nu (y). $$ Note: because of the asymmetry of the left self distributive law, the above order of integration is important (it is possible to show that there is an idempotent with the opposite order of integration). I conjecture a positive answer (this would follow from a positive answer for the free binary system on one generator) and this question may be easier to resolve than the other.

This is a follow up question to MO question "Idempotent measures on the free binary system?". Let $(A,*)$ be the free binary operation on one generator which satisfies the left self distributive law: $a * (b * c) = (a * b)* (a* c)$. Is there a finitely additive probability measure $\mu$ on $A$ such that $\mu * \mu = \mu$?

Here $\mu * \nu$ is defined by $$ \mu * \nu (X) = \int \int \mathbf{1}_{*^{-1}(X)} (x,y)\ d\mu (x) \ d \nu (y). $$ Note: because of the asymmetry of the left self distributive law, the above order of integration is important (it is possible to show that there is an idempotent with the opposite order of integration). I conjecture a positive answer (this would follow from a positive answer for the free binary system on one generator) and this question may be easier to resolve than the other.

This is follow up question to MO question "Idempotent measures on the free binary system?" Let $(A,*)$ be the free binary operation on one generator which satisfies the left self distributive law: $a \* (b \* c) = (a \* b)\* (a\* c)$. Is there a finitely additive probability $\mu$ on $A$ such that $\mu \* \mu = \mu$?

Here $\mu * \nu$ is defined by $$ \mu * \nu (X) = \int \int \mathbf{1}_{\*^{-1}(X)} (x,y)\ d\mu (x) \ d \nu (y). $$ Note: because of the asymmetry of the left self distributive law, the above order of integration is important (it is possible to show that there is an idempotent with the opposite order of integration). I conjecture a positive answer (this would follow from a positive answer for the free binary system on one generator) and this question may be easier to resolve than the other.

This is follow up question to MO question "Idempotent measures on the free binary system?" Let $(A,*)$ be the free binary operation on one generator which satisfies the left self distributive law: $a * (b * c) = (a * b)* (a* c)$. Is there a finitely additive probability $\mu$ on $A$ such that $\mu * \mu = \mu$?

Here $\mu * \nu$ is defined by $$ \mu * \nu (X) = \int \int \mathbf{1}_{*^{-1}(X)} (x,y)\ d\mu (x) \ d \nu (y). $$ Note: because of the asymmetry of the left self distributive law, the above order of integration is important (it is possible to show that there is an idempotent with the opposite order of integration). I conjecture a positive answer (this would follow from a positive answer for the free binary system on one generator) and this question may be easier to resolve than the other.