Let (Ω, Σ, µ) be a finite measure space and X, a Banach space with continuous dual X*. A scalarly measurable function f: Ω→X is Dunford integrable if for each x* X*, x*f L1(µ). Define the operator Tf. X* → L1(µ) by T(x*) = x*f. Then f is Pettis integrable if and only if this operator is weak*-to-weak continuous. This paper begins with an overview of this function. Work by Robert Huff and Gunnar Stefansson on the operator Tf motivates much of this paper. Conditions that make Tf weak*-to-weak continuous are generalized to weak*-to­weak continuous operators on dual spaces. For instance, if Tf is weakly compact and if there exists a separable subspace D X such that for each x* X*, x*f = x*fχDµ-a.e, then f is Pettis integrable. This nation is generalized to bounded operators T: X* → Y. To say that T is determined by D means that if x*| D = 0, then T (x*) = 0. Determining subspaces are used to help prove certain facts about operators on dual spaces. Attention is given to finding determining subspaces far a given T: X* → Y. The kernel of T and the adjoint T* of T are used to construct determining subspaces for T. For example, if T*(Y*) ∩ X is weak* dense in T*(Y*), then T is determined by T*(Y*) ∩ X. Also if ker(T) is weak* closed in X*, then the annihilator of ker(T) (in X) is the unique minimal determining subspace for T.