Let $d\in\mathbb N$, $f:\mathbb R^d\to\mathbb R$ be Lebesgue integrable and vanishing outside $[0,1)^d$. Moreover, let $k\in\mathbb N$ and $x_1,\ldots,x_k\in[0,1)^d$. Let $w_1,\ldots,w_k\ge0$ with $\sum_{i=1}^kw_i=1$ and suppose we want to approximate $$I:=\int f(x)\:{\rm d}x$$ by $$\tilde I:=\sum_{i=1}^kw_if(x_i).$$ If we write $$\sigma(x):=\sum_{i=1}^kw_i\delta(x-x_i)\;\;\;\text{for }x\in\mathbb R^d$$ and denote the Fourier transform of a function $g$ by $\hat g$, we see that $$\tilde I=\int\hat\sigma(\omega)\hat f(\omega)\:{\rm d}\omega.$$
My question is: Suppose $\hat\sigma$ vanishes in a punctured ball $B_r(0)\setminus\{0\}$, where $r>0$, and $\hat f$ vanishes outside the ball $B_r(0)$. Are we then able to show that the approximation is exact; i.e. $\tilde I=I$?
For this, note that we have $\hat\sigma(0)=\sum_{i=1}^kw_i=1$ and hence $$I=\hat f(0)=(\hat\sigma\hat f)(0)\tag1.$$