Prove $f$ is not Riemann Integrable using alternative definition of Riemann Integrability

Question

Alternative definition $f : [a,b] \to \mathbb{R}$ is Riemann-Integrable if $\forall \varepsilon > 0, \exists$ step functions $f_\varepsilon, g_\varepsilon$ such that $\lvert f- f_\varepsilon \rvert \le g_\varepsilon$ and $Ig_\varepsilon \le \varepsilon$ where $Ig_\varepsilon = \sum_{j = 0}^{N}g_\varepsilon(x)(\Delta x_{j+1} - \Delta x_j)$

$f : x \in [0,1] \mapsto 0$ if $x \in ([0,1]- \mathbb{Q})$ else $x \mapsto 1$

I know this isn't Riemann integrable and can show it using the upper and lower sums definition.

For this definition I am not entirely sure. I mean it seems like a very similar idea.

I thought perhaps the route to take is assume that is is Riemann-Integrable then let $\varepsilon < 1$ (as it should hold if it were in fact R-I). Now given any partition of the interval, each block will contain rational and irrational numbers so over the same block $\lvert f - f_\varepsilon| = 0$ &$1$ as I could take $f_\varepsilon := f$ as $f$ 'acts" like a step function.

Answer 1

You need to look at $\epsilon<1/2$. Then let $x_1,...,x_n$ be the partitioning of $f_\epsilon$. You can refine it to also be a partitioning of the $g_\epsilon$.

Inside of $[x_i,x_{i+1}]$ the function $f_\epsilon$ and $g_\epsilon$ take on a constant values $c_i,d_i$, so $|f-f_\epsilon|$ takes either the value $|1-c_i|$ or $|c_i|$ and both bust be smaller than $d_i$ (since both values are taken as long as $x_i\neq x_{i+1}$). Obviously either $|1-c_i|<1/2$ or $|c_i|<1/2$ but never both. So $d_i≥1/2$ and $g_\epsilon>1/2$ everywhere.

The contradiction $I(g_\epsilon)<\epsilon<\frac12$ now follows.