Riemann Integral of Piecewise Function

Question

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.

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 $1$.