# 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 =)

1. 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).
2. 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 $$\varepsilon > 0$$.