Since $f$ assumes the value $0$ except that at a single point $x\in I$, $f(x)$ is non-zero.
And this point $x$ falls in one generalized rectangle in the partition.
The fineness of partition is important because the size of the rectangle has measure controlled by this.
Then, In order to keep the fineness of partition, let choose small enough volume $V_p$ of the relevant part, which contains of $x$.
And let's assume that $f$ is positive.
Then, the lower sum is zero since the value of $f$ is zero except at $x$.
i.e $L(f,P)=0$
But, the upper sum has just one nonzero contribution. And the contribution is at most $f(x)\cdot V_p$
So, $U(f,P)=f(x)\cdot V_p$
Let choose $\epsilon >0$ such that $V_p< \frac{\epsilon}{f(x)}$
Then, $$U(f,P)-L(f,P)=f(x) V_p -0< f(x) \frac{\epsilon}{f(x)}=\epsilon$$
Thus, f is integrable.
As for $\int_I f$,
as the partition $P$ becomes finer and the upper and lower sums converge to zero, the integral of $f$ is zero.
Note that this answer is the combination of N. R. Peterson's answer and M. Bennet's good and wide explanation.