I am trying to solve the following problem:
Let $(X, \mathcal{M},\mu)$ a probability space, $\varphi:\mathbb{C} \to \mathbb{R}$ a convex function and $f \in \mathcal{L}^1(\mu,\mathbb{C})$ such that $\varphi \circ f \in \mathcal{L}^1(\mu,\mathbb{R})$. Let $\Sigma \subseteq \mathcal{M}$ be a $\sigma$-algebra and let $g=\mathbb{E}(f,\Sigma)$, $h=\mathbb{E}(\varphi \circ f,\Sigma)$, where $\mathbb{E}(-,\Sigma)$ denotes the conditional expectation with respect to $\Sigma$. Prove that for every $\lambda \in \mathbb{R}$ $$ \mu (\left \lbrace x \in X : \varphi \circ g (x) > \lambda \geq h(x) \right \rbrace)=0. $$
and I am also given the following hint
If $B$ is a measurable set for $\Sigma$, contained in $\left \lbrace x \in X :\lambda \geq h(x) \right \rbrace$ and $\mu(B)>0$, then $$ \varphi \left( \frac{1}{\mu(B)}\int_Bg d\mu \right) \leq \lambda. $$
However, when one tries to use the hint for $B=\left \lbrace x \in X : \varphi \circ g (x) > \lambda \geq h(x) \right \rbrace$ and reasoning by contradiction, we only get
$$ \varphi \left( \frac{1}{\mu(B)}\int_Bg d\mu \right) \leq \lambda < \frac{1}{\mu(B)} \int_B (\varphi \circ g) d\mu.$$ I would really appreciate any suggestion for a continuation leaving to a contradiction or other ways to tackle this problem.