integral computed with respect to a sub-$\sigma$-algebra Let $\mathcal M_0$ be a $\sigma$-algebra that is contained in a $\sigma$-algebra $\mathcal M$ of subsets of a set $X$, $\mu$ a measure on $\mathcal M$ and $\mu_0$ the restriction of $\mu$ to $\mathcal M_0$. Let $f$ be a nonnegative real-valued function that is measurable with respect to $\mathcal M_0$ (and hence with respect to $\mathcal M$ as well). Then the set of all nonnegative simple functions $\phi\le f$ measurable with respect to $\mathcal M_0$ is a subset of the set of all nonnegative simple functions $\psi\le f$ measurable with respect to $\mathcal M$, and so
$$
\int_Xf\,d\mu_0\le\int_Xf\,d\mu.
$$
Can this inequality be strict?
 A: The intuition here is that integrating $\mathcal M_0$-measuarble function w.r.t. $\mu_0$ is actually integrating it w.r.t. $\mu$ since you are doing everything on the restriction where both measures agree. Let me provide a formal way of proving this: consider "two" measurable spaces $(X,\mathcal M)$ and $(X,\mathcal M_0)$ - this trick is often used in topology to study different typologies on the same set. You have that
$$
  \mathrm{id}:X\to X \quad\text{ given by }\quad \mathrm{id}(x) = x \;\forall x\in X \tag{1}
$$
is a $\mathcal M/\mathcal M_0$-measurable map, and so induces the image measure $\mathrm{id}_*\mu$ on $(X,\mathcal M_0)$ define by
$$
  (\mathrm{id}_*\mu)(A) = \mu(\mathrm{id}^{-1}(A)) = \mu(A) \qquad\forall A\in \mathcal M_0. \tag{2}
$$ 
From $(2)$ it is clear that we can represent $\mu_0 = \mathrm{id}_*\mu$, so that for any non-negative $f\in \mathcal M_0$:
$$
\begin{align}
  \int f\,\mathrm d\mu_0 &= \int f\,\mathrm d(\mathrm{id}_*\mu) &\text{from }(2)
  \\
  &= \int (f\circ \mathrm{id})\,\mathrm d\mu  &\text{change of variables}
  \\
  &= \int f\mathrm d\mu  &\text{from }(1)
\end{align}
$$
where we used the change of variables formula. 
Perhaps, there may be a more basic proof - but I believe it is important to learn techniques of dealing with different structures on the same set using the $\mathrm{id}$ map.
A: No, you have equality.
This is because you can find a nondecreasing sequence of nonnegative, $\mathcal M_0$-measurable functions $(\phi_n)$ such that $\phi_n(x)\to f(x)$ for all $x\in X$. By the monotone convergence theorem applied to $\mu$ and $\mu_0$ it follows that $\int fd\mu=\int fd\mu_0$.
