Let $(X,\mathcal{E},\mu)$ denote a measure space. Then each measureable function $f : X \rightarrow [0,\infty]$ has a Lebesgue integral $\int f d\mu$ given by take a supremum of the integrals of simple functions that are bounded above by $f$.
Question. Suppose we instead try taking infima of simple functions that are bounded below by $f$. I assume the resulting integral is "poorly behaved", since no one ever talks about it. My question is quite simply, what happens if we do this?
Consider the function $f(x)= x^{-\frac12}$ on $(0,1)$. Note that the function values can be arbitrarily large.
Since a simple function takes only finitely many values, every simple function that is bounded below by $f$ has to be infinite on a set of non-zero measure.
Thus, the integral using your suggested infimum definition would be $\infty$, whereas the usual Lebesgue integral would have a finite value.
This is just an example, but it demonstrates the difficulty that will arise with non-bounded functions. For bounded functions $f$ your infimum definition is equivalent to the usual defintion, see the comment of Crostul.