Recall the two integrals used to calculate volume along a ray in a NeRF:

\[c = \int_0^\infty T(t) \sigma(t) c(t) \,dt \qquad \qquad T(t) = \exp\left(-\int_0^t \sigma(s) \,ds\right)\]

In the first integral, \(T(t)\sigma(t)\) are continuous weights along the ray representing how much of the color \(c(t)\) will reach the camera through the volume that exists in between. It would make sense for these weights to satisfy the following relation for any \(\tau\):

\[\int_0^\tau T(t) \sigma(t) \,dt \in [0, 1]\]

Namely with the integral taking the value of 0 if the volume between \(0\) and \(\tau\) is totally transparent and 1 if the volume between \(0\) and \(\tau\) is totally opaque. In our previous derivations we can see that \(T, \sigma \geq 0\) by definition. This gives us our lower bound trivially. Furthermore, we can see that \(T \in [0, 1]\) trivially from definition. In our previous derivations, we show that \(dT(\tau) = -\sigma(\tau) T(\tau)\) and therefore our upper bound arises as a consequence of the fundamental theorem of calculus:

\[\int_0^\tau T(t) \sigma(t) d,dt = T(0) - T(\tau) = 1 - T(\tau) \leq 1\]