Derivation of Casher-Banks relation Consider two-point function $\langle \bar{\psi}\psi\rangle$ in a model with massive fermions $\psi$ and gauge field:
$$
\langle \bar{\psi}\psi\rangle =\frac{1}{V}\sum_{n} \frac{1}{\lambda_{n} +im}, 
$$
where $\lambda_{n}$ is the eigenvalue of the corresponding Dirac operator (with $\lambda_{0} = 0$). The expression can be rewritten in the form
$$
\tag 1 \langle \bar{\psi}\psi\rangle = \int d\nu \rho(\nu)\frac{1}{\nu + im},
$$
where $\rho(\nu) = \frac{1}{V}\sum_{n}\delta(\nu - \lambda_{n})$.
On p. 56 of the paper is stated that by applying thermodynamic limit and second the limit $m \to 0$ to $(1)$, we'll obtain zero; this sounds reasonable since we have then finite system where spontaneous symmetry breaking corresponding to non-zero order parameter can't occur. However, some small (although general) detain is unclear for me, and I don't understand why numerically it is so.
Can anyone please explain it for me?
 A: Consider a theory with chiral symmetry breaking condensate $\Sigma$. If we study the spectrum of the euclidean Dirac operator in a finite volume $V$, then on average the first non-zero eigenvalue is located at $\lambda\sim 1/(V\Sigma)$, and a finite density of eigenvalues at $\lambda=0$ only emerges in the thermodynamic limit $V\to\infty$. This has been studied in great detail in random matrix theory, see the survey article. 
Postscript: Consider a euclidean volume $V$ with $N_+$ instantons and $N_-$ anti-instantons. The topological charge is $Q=N_+-N_-$, and by the index theorem the number of exact zero modes is $Q$. The remaining $N_++N_--|Q|$ instanton zero modes get lifted and become small eigenvalues. Because of the fermion determinant the probability of encountering $Q$ eaxct zero modes scales as $m^{N_f|Q|}$, so the contribution to the spectral density is at most $\rho(\lambda)\sim m^{N_F}\delta(\lambda)$. This means that for $N_f>1$ the chiral condensate is not related to exact zero modes, but to the quasi-zero modes. 
A: I am not completely sure about what you are asking, but may be, this is useful -- 
Using the result,  
\begin{equation} 
\displaystyle\lim_{m \to 0} \frac{1}{m + i \lambda} = \pi \delta(\lambda) - i \mathcal{P} (\frac{1}{\lambda}) 
\end{equation}
where $\mathcal{P}$ is the principal value, we can show that,  
\begin{equation}
 \langle \overline{\psi} \psi \rangle  = \int_{-\infty}^{+\infty} d\lambda ~ \rho(\lambda) \frac{1}{m + i \lambda} = \pi \rho(0) 
\end{equation}
