Suppose $f$ is Riemann integrable over $[a,b]$. The proof I am reading that shows that $f$ must be bounded, just says if $f$ is unbounded then there is some point $x$, such that for any $n\in\mathbb{N}$ and $\epsilon>0$ if we take any partition divded into $n$ parts we have $|f(x)|>n\epsilon$ and so if we take $x$ as a tag in the partition then $$|S(f,\mathcal{P})-L|>\epsilon\space\space\space\space\space\space (\dagger),$$ where $S(f,\mathcal{P})$ is the Riemann sum of $f$ with respect to the partition $\mathcal{P}$ and $L=\int_a^bf$.
My trouble is trying to get $(\dagger)$, I am not sure how split up the Riemann sum to get the inequality, I was thinking the reverse triangle but get nowhere. So any help will be really appreciated and needed.
Thanks in advance
Best Answer
You have not told us what your working definition of Riemann integrability is.
If integrability is defined in terms of lower and upper sums, and $f$ is unbounded above, then every single upper sum is undefined (or $=\infty$, if you prefer). Basta; no further epsilontics needed.
If integrability is defined in terms of quantities like $\|\Delta f\|_{I_j}:=\sup_{x, \ y\in I_j}|f(x)-f(y)|$ then again for every partition at least one of these quantities is $=\infty$, hence you can never make $\sum_j\|\Delta f\|_{I_j}\>|I_j|<\epsilon$. But this is required before you can even think of computing "general Riemann sums" $\sum_{j=1}^N f(\tau_j)\>|I_j|$ as approximations to the intended integral.