ADD It seems you're being given Darboux's approach to integration. I guess that for each partition $P=\{a=x_0,\dots,x_n=b\}$ of the interval $[a,b]$ you're defining
$$M_k=\sup_{x\in[x_{k-1},x_{k}]}f(x)$$
$$m_k=\inf_{x\in[x_{k-1},x_{k}]}f(x)$$
and then the lower and upper sums of $f$ over $P$ as $$U(f,P)=\sum_{k=1}^n M_k(x_k-x_{k-1})$$
$$L(f,P)=\sum_{k=1}^n m_k(x_k-x_{k-1})$$
and the the lower and upper integrals (which always exist) as
$$\overline{\int_a^b}f=\inf\{U(f,P):P\text{ is a partition of }[a,b]\}$$
$$\underline{\int_a^b}f=\sup\{L(f,P):P\text{ is a partition of }[a,b]\}$$
The prove a function is not Riemann (equiv. Darboux) integrable, you can show those last number differ. But you have to try many partitions to see what is really going on, since the supremum and infimum are taken when $P$ varies throughout all possible partitoins of $[a,b]$. In particular, $M_k$ and $m_k$ will usualy vary for different partitions, as the comments show.
I hope you can see that $$\underline{\int_{-2}^3} f=0$$
Now you have to prove that $$\overline{\int_{-2}^3} f$$ is bounded away from zero, and you'll have proven the integral cannot exist.
To prove both assertions, use that both the irrationals and rationals are dense in $\Bbb R$. For each partition $P=\{-3=x_1,x_2,\dots,x_{n-1},x_n=2\}$, can you see why
$$\inf_{[x_{k-1},x_k]}f(x)=0$$
for any interval in the partition, for example?
On the other hand, what is the minimum value $2|x|+1$ takes on $[-2,3]$? What does this tell you, plus the density of $\Bbb Q$ on $\Bbb R$ about $$\overline{\int_{-2}^3} f\text{ ? }$$
Best Answer
Set of points of discontinuity is not measure zero