Prove that the lower Riemann integral of $f$ on $[a, b]$ Prove that the lower Riemann integral of $f$ on $[a, b]$ is $\sup\{\int_a^b\psi: \psi \text{ is a step function on } [a, b],\text{ such that } \psi\leq f \text{ on }[a, b]\}$.
The definitions that I am working with are:
For a partition $P = \{x_0, ..., x_n\}$ of $[a,b]$ with $a = x_0 < x_1 < ... < x_n = b$. Then
$\int_a^b\psi = \sum_{j=1}^{n}\psi_{j}(x_j - x_{j-1})$ where $\psi_i$ is the constant value of $\psi$ on the interval $\psi$ on $[x_{j-1}, x_j]$.
The Riemann integral of $f$ is defined to be (using the partition $P$ above)
$\int_a^b f = L(f,[a,b]) = U(f,[a,b])$ where
$L(f,[a,b]) = \sum_{j=1}^{n}(x_j-x_{j-1})\inf_{[x_{j-1},x_j]}f~~,
U(f,[a,b]) = \sum_{j=1}^{n}(x_j-x_{j-1})\sup_{[x_{j-1},x_j]}f
 ~$are the lower and upper Riemann sums respectively.
I was trying to show that the supremum over the step functions has to be equal to the lower sum/ the constant value $\psi_j = \inf f$ on any of our sub intervals, but going from here to proving that the sums are equal I was running to to troubles keeping my epsilons straight.
Basically, I wanted to show that for our step functions we need that the step function value needs to be equal to the $\inf f$ on any of the sub intervals and from there we'd hopefully get the desired equality.
Any help on if this is the correct direction to go/ what a clean proof looks like is appreciated. Thanks =)
 A: You want to show two things:

*

*For every step function $\psi$ with $\psi \le f$ we have $\int \psi \le \underline{\int} f$. And to do this, simply define a partition so that $\psi$ is constant on each interval $(x_i, x_{i+1})$ (we don't need to worry about the endpoints of the intervals but do check this).


*For every $\varepsilon$, we can find a step function $\psi$ such that $\underline{\int} f \le \int \psi + \varepsilon$. And to do this, you can do exactly as you say: define $\psi_i = \inf_{[x_i, x_{i+1}]} f(x)$ and define $\psi$ piecewise using the constants $\psi_i$.
For 1. you want to check that $\int \psi \le L(f, \mathcal{P})$ for any sufficiently small partition (refining the partition given by $\psi$). It follows then that $\int \psi \le \sup_{\mathcal{P}} L(f, \mathcal{P}) = \underline{\int} f$.
For 2. you can check that $\int \psi = L(f, \mathcal{P})$ by definition and by definition of $\underline{\int} f = \sup_{\mathcal P} L(f, \mathcal{P})$ there exists a partition $\mathcal{P}$ such that $\underline{\int} f \le L(f, \mathcal{P}) + \varepsilon$ (that's the definition of $\sup$).

This is a common strategy: if you want to show $A = B$ you can show that $A \le B$ and $B \le A + \varepsilon$ for every $\varepsilon > 0$.
