I am only concerned with the integral of a measurable function $f : M \to \mathbb{R}^{\ast}_{+}$, where $\mathbb{R}^{\ast}_{+}$ contains all non-negative real numbers and positive infinity. As I understand it, this is defined as $$\sup \left\{ \int s : \text{$s$ is simple and $s\le f$}\right\}.$$
My question is whether the integral is also equal to $$ \inf \left\{ \int s : \text{$s$ is simple and $s\ge f$}\right\}.$$
On
If said function is measurable then yes, it is. Some textbooks say that a function $f$ is Lebesgue-integrable if and only if $\sup\{s \text{ simple} : s \leq f\} = \inf\{s \text{ simple} : s \geq f\}$.
It all comes down to showing that these definitions of measurable function $f:X \rightarrow \mathbb{R}$ are all equivalent:
$\forall a \in \mathbb{R}$ the set $\{x \in X : f(x) < a\}$ is measurable.
$\forall a \in \mathbb{R}$ the set $\{x \in X : f(x) \leq a\}$ is measurable.
$\forall a \in \mathbb{R}$ the set $\{x \in X : f(x) > a\}$ is measurable.
$\forall a \in \mathbb{R}$ the set $\{x \in X : f(x) \geq a\}$ is measurable.
which is a piece of cake to prove using the property of closure under complement of a sigma algebra.
No.
Let $f:(0,1]\to[0,+\infty]$ be prescribed by $x\mapsto x^{-\frac12}$
Then $\int fd\lambda=\left[2x^{\frac12}\right]^1_0=2$ and this is the outcome of the supremum in your question.
However if $s$ is a simple function then its image is finite so that we can find a value $y>0$ such that $s(x)<y$ for all $x\in(0,1]$. But we can also find a value $x_0$ with $f(x_0)=y$.
Conclusion: no simple function $s$ exists that satisfies $f\leq s$.
Then the infimum in your question is the infimum over the empty set hence takes value $+\infty$.