So I have this function and want to find out what the Riemann integral between 0 and 1 inclusive is, I have already proved that the upper Riemann integral is 1 but am having trouble formalising why the lower Riemann integral is 1 to complete the proof:

$g(x) = 1$ if $x\neq0$ else $g(x) = 0$

So the Lower Riemann integral is the supremum of all possible lower Riemann sums,

My logic is as follows:

Considering any partition the infimum on each interval except the first will always be 1, and the infimum on the first interval $[0,R]$ where R is a positive real will always be zero

Let P be any partition of $[0,1]$

$L(g,P) = \sum_{k=1}^nm_k|I_k|$ where $m_k$ is the infimum on of $f(x)$ on $I_k$

Clearly this means the first term is zero and we arrive at

$L(g,P) = 1- |I_1|$

Now clearly 1 is an upperbound but how do i formalise that 1 is the supremum of this set of Riemann sums, thankyou.

  • $\begingroup$ That's not true, the function is zero only at zero and one for the rest of the real line. Edit: user deleted the comment to which this a response to. $\endgroup$ Oct 16, 2019 at 17:38
  • $\begingroup$ Consider the sequence of partitions $\mathcal P_n = \{0\}\cup\{1/k:1\leqslant k\leqslant n\}$. Then the lower Riemann sum of $\mathcal P_n$ is $1/n$ which tends to zero. $\endgroup$
    – Math1000
    Oct 16, 2019 at 17:39

1 Answer 1


Let $\varepsilon > 0$ be given. Then a possible lower Darboux sum is $$L_\varepsilon=0\varepsilon+1(1-\varepsilon)=1-\varepsilon$$ This means that the supremum of all of the possible lower sums, i.e. $$\underline{\int} f \geqslant 1$$ But we also have that $$\overline{\int}f \geqslant \underline{\int} f$$ $$\implies 1 \leqslant \underline{\int} f \leqslant 1$$ $$\implies \underline{\int}f=1$$ So $$\overline{\int}f = \underline{\int} f=1$$ So $f$ is Darboux integrable with integral $1$.


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