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.


1 Answer 1


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 $2\delta\sum_{k=1}^nm_k^* \leqslant 2nM\delta$.


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

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

  • $\begingroup$ 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. $\endgroup$
    – xyz
    Jun 24, 2020 at 3:41
  • $\begingroup$ @mathodfun: You’re welcome. I’m glad I could help. $\endgroup$
    – RRL
    Jun 24, 2020 at 4:16

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