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There is a conditional version of the Jensen inequality that may be what you are looking for. $$ f(E(X\mid {\cal F})) \leq E(f(X) \mid {\cal F}) $$ Taking the expectation, this gives $$ E(f(E(X\mid {\cal F}))) \leq E(f(X)). $$ So you can replace the expectation $E(X) = \int X \, d\mu$ by any averages $E(X\mid {\cal F})$ along the atoms of a $\sigma$-algebra ${\cal F}$. Assuming that there are conditional probabilities associated to ${\cal F}$, we get $E(X\mid {\cal F})(\omega) = \int X(\omega) \, d\mu(\omega \mid {\cal F})$.

The simplest case is given by a $\sigma$-algebra associated to a countable partition $\{E_i\}_{i\in {\bf N}}$ of the underlying space. This gives $$ \int f\Bigl( \sum_i {1\over \mu(E_i)} \int_{E_i} X \, d\mu \ {\bf 1}_{E_i} \Bigr) \, d\mu\leq \int f(X) \, d\mu. $$

There is a conditional version of the Jensen inequality that may be what you are looking for. $$ f(E(X\mid {\cal F})) \leq E(f(X) \mid {\cal F}) $$ Taking the expectation, this gives $$ E(f(E(X\mid {\cal F}))) \leq E(f(X)). $$ So you can replace the expectation $E(X) = \int X \, d\mu$ by any averages $E(X\mid {\cal F})$ along the atoms of a $\sigma$-algebra ${\cal F}$. Assuming that there are conditional probabilities associated to ${\cal F}$, we get $E(X\mid {\cal F})(\omega) = \int X(\omega) \, d\mu(\omega \mid {\cal F})$.

The simplest case is given by a $\sigma$-algebra associated to a countable partition $\{E_i\}_{i\in {\bf N}}$ of the underlying space. This gives $$ \sum_i {\mu(E_i)} \ f\Bigl({1\over \mu(E_i)}\int_{E_i} X \, d\mu\Bigr) \leq \int f(X) \, d\mu. $$

There is a conditional version of the Jensen inequality that may be what you are looking for. $$ f(E(X\mid {\cal F})) \leq E(f(X) \mid {\cal F}) $$ Taking the expectation, this gives $$ E(f(E(X\mid {\cal F}))) \leq E(f(X)). $$ So you can replace the expectation $E(X) = \int X \, d\mu$ by any averages $E(X\mid {\cal F})$ along the atoms of a $\sigma$-algebra ${\cal F}$. Assuming that there are conditional probabilities associated to ${\cal F}$, we get $E(X\mid {\cal F})(\omega) = \int X(\omega) \, d\mu(\omega \mid {\cal F})$.

There is a conditional version of the Jensen inequality that may be what you are looking for. $$ f(E(X\mid {\cal F})) \leq E(f(X) \mid {\cal F}) $$ Taking the expectation, this gives $$ E(f(E(X\mid {\cal F}))) \leq E(f(X)). $$ So you can replace the expectation $E(X) = \int X \, d\mu$ by any averages $E(X\mid {\cal F})$ along the atoms of a $\sigma$-algebra ${\cal F}$. Assuming that there are conditional probabilities associated to ${\cal F}$, we get $E(X\mid {\cal F})(\omega) = \int X(\omega) \, d\mu(\omega \mid {\cal F})$.

The simplest case is given by a $\sigma$-algebra associated to a countable partition $\{E_i\}_{i\in {\bf N}}$ of the underlying space. This gives $$ \int f\Bigl( \sum_i {1\over \mu(E_i)} \int_{E_i} X \, d\mu \ {\bf 1}_{E_i} \Bigr) \, d\mu\leq \int f(X) \, d\mu. $$

There is a conditional version of the Jensen inequality that may be what you are looking for. $$ f(E(X\mid {\cal F})) \leq E(f(X) \mid {\cal F}) $$ Taking the expectation, this gives $$ E(f(E(X\mid {\cal F}))) \leq E(f(X)).$$

There is a conditional version of the Jensen inequality that may be what you are looking for. $$ f(E(X\mid {\cal F})) \leq E(f(X) \mid {\cal F}) $$ Taking the expectation, this gives $$ E(f(E(X\mid {\cal F}))) \leq E(f(X)). $$ So you can replace the expectation $E(X) = \int X \, d\mu$ by any averages $E(X\mid {\cal F})$ along the atoms of a $\sigma$-algebra ${\cal F}$. Assuming that there are conditional probabilities associated to ${\cal F}$, we get $E(X\mid {\cal F})(\omega) = \int X(\omega) \, d\mu(\omega \mid {\cal F})$.