# Equivalence definition of Lower Integral

let $$f$$ be a real-valued bounded function on $$[a,b]$$.

For all partition $$P:x_0,...x_N$$ of $$[a,b]$$, define $$m_k(f,P)=\inf_{x_{i-1}\le x\le x_i}(f(x))$$ and $$m^*_k(f,P)=\inf_{x_{i-1}< x< x_i}(f(x))$$ (which does not include the endpoints) for all $$k=1,...,N$$.

The lower integral of $$f$$ is usually defined by $$L=\sup\{\sum_{k=1}^Nm_k(f,P)(x_i-x_{i-1}): P:x_0,...,x_N$$ is a partition of $$[a,b]\}$$. But intuitively, the value of the lower integral should remains the same if we replace $$m$$ by $$m^*$$, i.e $$L=L^*:=\sup\{\sum_{k=1}^Nm^*_k(f,P)(x_i-x_{i-1}): P:x_0,...,x_N$$ is a partition of $$[a,b]\}$$, since only 2 points are removed for each section and should not affect the whole integral.

It is clear that $$L\le L^*$$, since each $$m_k(f,P)\le m_k^*(f,P)$$ by the property of infimum, but I am stuck in showing the another direction of the equality.

For a given partition $$P = (x_0,x_1, \ldots x_n)$$, let $$L(f,P) = \sum_{k=1}^nm_k(x_k - x_{k-1})$$ denote the usual lower Darboux sum and let $$L^*(f,P) = \sum_{k=1}^nm_k^*(x_k - x_{k-1})$$ denote the lower sum with infima taken over open subintervals.

You already have shown that $$L(f,P) \leqslant L^*(f,P)$$ which implies that

$$L = \sup_P L(f,P) \leqslant \sup_PL^*(f,P) = L^*$$

To prove that $$L = L^*$$, it is enough to show that for any $$\epsilon >0$$ there exists a partition $$Q$$ such that $$L^* - L(f,Q) < \epsilon$$.

Since $$f$$ is bounded, we have $$m < f(x) < M$$ for all $$x \in [a,b]$$. Also, for any $$\epsilon > 0$$, there exists a partition $$P = (x_0,x_1,\ldots, x_n)$$ such that $$L^* - L^*(f,P) < \frac{\epsilon}{2}$$ (since $$L^* = \sup_PL^*(f,P)$$).

Define the partition $$Q = (x_0, x_0+\delta, x_1-\delta, x_1,x_1+\delta,\ldots, x_n-\delta,x_n)$$ where

$$0 < \delta < \min\left(\frac{\max_{1\leqslant j \leqslant n}(x_j - x_{j-1})}{2}, \frac{\epsilon}{4n(M-m)}\right)$$

We have

$$L(f,Q) = \sum_{k=1}^n\left(\inf_{x \in [x_{k-1},x_{k-1} + \delta]}f(x)\cdot\delta + \inf_{x \in [x_{k-1}+ \delta,x_{k} - \delta]}f(x)\cdot (x_k - x_{k-1} - 2\delta)+ \inf_{x \in [x_{k}-\delta,x_{k} ]}f(x) \cdot\delta\right)$$

Since $$\inf_{x \in [x_{k-1},x_{k-1} + \delta]}f(x), \, \,\inf_{x \in [x_{k}-\delta,x_{k}]}f(x) \geqslant m$$ and $$\inf_{x \in [x_{k-1}+ \delta,x_{k} - \delta]}f(x) \geqslant m_k^*$$ it follows that

$$L(f,Q) \geqslant \sum_{k=1}^n\left(m\cdot\delta + m_k^*\cdot (x_k - x_{k-1} - 2\delta)+ m \cdot\delta\right) \\ = \sum_{k=1}^nm_k^*\cdot (x_k - x_{k-1}) - 2\delta\sum_{k=1}^nm_k^* + 2nm\delta$$

The first sum on the RHS is just $$L^*(f,P)$$ and for the second sum we have $$2\delta\sum_{k=1}^nm_k^* \leqslant 2nM\delta$$.

Thus,

$$L(f,Q) \geqslant L^*(f,P) - 2n(M-m)\delta > L^* - \frac{\epsilon}{2} - 2n(M-m) \frac{\epsilon}{4n(M-m)}= L^*- \epsilon$$

Therefore, $$L = \sup_P L(f,P) = L^*$$ since for any $$\epsilon > 0$$ there exists a partition $$Q$$ such that $$L^* - L(f,Q) < \epsilon$$.

• Thank you very much for your help. Not only for this question, but other question you answered in the past, it really helped me out.
– xyz
Jun 24, 2020 at 3:41
• @mathodfun: You’re welcome. I’m glad I could help.
– RRL
Jun 24, 2020 at 4:16