Homological class of a singular variety Suppose $X$ is a compact complex manifold and $V\subseteq X$ is an irreducible analytic variety. Since $V$ may not be smooth, how does it make sense of saying $[V]$ as a homological class?
In the case when $V$ is a smooth $k$-dimensional submanifold, its Poincare dual $\eta_V\in H^{n-k}(X)$ can be thought of:


*

*The unique class in $H^{n-k}(X)$ such that $\int_V\omega=\int_X \omega\wedge\eta_V$ for all $\omega\in H^k(X)$

*The unique class in $H^{n-k}(X)$ such that $\int_W\eta_V=\text{Int}(V,W)$ for all $(n-k)$-dimensional submanifold, where Int is the intersection number.
How to interpret those in the case when $V$ is singular? Is it true that the singular point of $V$ is measure zero so the integration makes sense?
 A: A highbrow way to define the fundamental class $[V]$ is to apply the desingularization theorem of Hironaka, which implies that there is a compact complex manifold $V'$ and a holomorphic map $\tau:V'\to V$ which is an isomorphism over the smooth locus $V^{sm}$. 
Let $i:V\hookrightarrow M$ be the inclusion and denote $k=\dim_{\mathbb C}V$. Take $[V']\in H_{2k}(V',\mathbb Z)$ as the fundamental class on the compact manifold, one defines the fundamental class $[V]$ as the pushforward via composite $(i\circ\tau)_*[V']\in H_{2k}(M,\mathbb Z)$.
Certainly there is still a Poincare pairing formulation in this setting, except that if one prefers de Rham cohomology, the pairing $[\omega]\mapsto \langle{[V]},[\omega]\rangle$ should be interpreted as integration 
$$\omega\mapsto \int_{V^{sm}}\omega$$
over the smooth locus $V^{sm}\subset V$.
One can show that the integral is convergent and vanish on exact forms, so is well defined on cohomology class. Therefore, it defines a linear functional on $H^{2k}(M)$, so there is Poincare dual class $\eta_V\in H^{2n-2k}(M)$.
More literature can be found in 
(1) Griffiths & Harris, Principles of Algebraic Geometry, Chapter 0, section 2 and section 4; 
(2) Voisin, Hodge Theory and Complex Algebraic Geometry, I, Chapter 11.
