Does such a function exist? If yes, it must be a very pathological case. I am talking here about Lebesgue integrability.
For instance, if $f(x)=1$ if $x$ is rational and zero otherwise, then $\int_0^1 f(x)dx = 0$. So you need to find an example more pathological than that. A possible example is the following.
Let $f(x)$ be the realization of a Gaussian random variable $Z_x$ with mean equal to $0$ and variance equal to $1$. Let us assume that the $Z_x$'s are identically and independently distributed. Such a function $f(x)$ is nowhere continuous, and can be seen as the realization of a white noise. However, you could argue that its integral on $[0,t]$ is the value $B(t)$ of a realization of a Brownian motion starting with $B(0)=0$, and measured at time $t$. Thus $\int_0^1 f(x) dx = B(1)$. Note that Brownian motions are nowhere differentiable, so maybe there is a contradiction in what I am saying here.
Anyway, I've never found counter-examples: a function bounded on $[0, 1]$ but not integrable in that interval. Can you show an example?
Let $f$ be a bounded function on $[0,1]$.
Either $f$ is measurable, and then $$ \int_0^1 |f| ≤ \sup |f|\ \int_0^1 1\,\mathrm d x = \sup |f| < \infty $$ so $f$ is integrable.
Either $f$ is not measurable. This exists iff you assume assume the axiom of choice. You can then take any non measurable set $\Omega$ and take $f = \chi_\Omega$ the characteristic function of this set, as suggested by Nate Eldredge. Then by definition, this function is not integrable.