About Vanishing of BRST commutator in path integral In Witten's paper Topological Quantum Field Theory, about formula (3.2), the property $\langle\{Q,\mathcal{O}\}\rangle=0$ depends on the assertion that,
$$Z_{\varepsilon}(\mathcal{O})= \int \mathcal{D}X \exp(\varepsilon Q) \left[\exp\left(-\frac{\mathcal{L'}}{e^2}\right)\mathcal{O} \right]$$
is independent on $\varepsilon$. Where does the assertion come from?
 A: Witten clearly writes the justification just on the line above the equation (3.2): the integral is independent because the integration measure is invariant under supersymmetry – the symmetry generated by $Q$.
Just to be sure, $Q$ is the infinitesimal generator which is why $\exp(\varepsilon Q)$ is a finite transformation generated by this generator: $\varepsilon$ is the argument ("Grassmann angle") of the transformation. And $\exp(\varepsilon Q) [{\mathcal A}]$ is the transformed operator ${\mathcal A}$ by this transformation and the integral is a device that produces a scalar out of a function of $X$. 
The independence on $\varepsilon$ holds because the SUSY-transformed integral of the SUSY-transformed (operator-valued) function is the same thing as the original integral of the SUSY-transformed function: I could have erased the adjective "SUSY-transformed" in front of the integral because the integration is SUSY-invariant. Because it doesn't matter whether we transform the integrand by $\exp(\varepsilon Q)$, it's the same thing as saying that the integral is independent of $\epsilon$ because it has the same value as the value for $\varepsilon=0$.
