Direct derivation of the Veneziano-Yankielowicz superpotential from matrix model

We derive the Veneziano-Yankielowicz superpotential directly from the matrix model by fixing the measure precisely. The essential requirement here is that the effective superpotential of the matrix model corresponding to the ${\cal N}=4$ supersymmetric Yang-Mills theory vanishes except for the tree gauge kinetic term. Thus we clarify the reason why the matrix model reproduces the Veneziano-Yankielowicz superpotential correctly in the Dijkgraaf-Vafa theory.