Is it, in general, true that $\mu_f \leq \mu$, where $\mu_f, \mu$ are defined in the following way... Let $(X, \mathcal{M}, \mu)$ be a measurable space and $f : X \to [0, \infty]$ a measurable function. Let 
$$\mu_f : \mathcal{M} \to [0, \infty], \mu_f(A) = \int_A f d\mu.$$
Is it, in general, true that 
$$\mu_f \leq \mu?$$
$$\mu_f \leq \mu \Leftrightarrow \mu_f(A) \leq \mu(A), \forall A \in \mathcal{M} \Leftrightarrow \int_A f d\mu \leq \mu(A), \forall A \in \mathcal{M} \Leftrightarrow \int_X f \cdot 1_A d\mu \leq \int_X 1_A d\mu, \forall A \in \mathcal{M}. $$
Can someone, please, give me a hint on how to show that $\mu_f \leq \mu$ or not?

As @user251257 pointed out in his comment, the claim does in general not hold true. In fact, one can show the following result

$\mu_f \leq \mu$ if, and only if, $\mu(f > 1) =0$.

Proof: 


*

*If $f \leq 1$ $\mu$-almost everywhere, then $$\mu_f(A) = \int_A \underbrace{f}_{\leq 1} \, d\mu \leq \int_A \, d\mu = \mu(A)$$ for any measurable set $A$. 

*Conversely, assume that $\mu(f > 1) > 0$. By the continuity of the measure $\mu$, we can find $n \in \mathbb{N}$ such that $A_n := \{f>1+1/n\}$ satisfies $\mu(A_n)>0$. Then $$\mu_f(A_n) = \int_{A_n} \underbrace{f}_{\geq 1+1/n} \, d\mu \geq \mu(A_n) \left(1+ \frac{1}{n} \right)>\mu(A_n),$$ i.e. $\mu_f \leq \mu$ does not hold true.

