Does a matched filter give a gain in SNR?
Output SNR for matched filter can be analysed in time domain, which is as explained below,

Consider a filter with Impulse response  $ \enspace h(t)$  and input $\,s(t)+w(t)$, where $\,s(t)$ represents the required signal and $\,w(t)$ represents AWGN noise. 

Output signal is given by  $$y(t) = s(t) * h(t) + w(t) * h(t) $$

Writing signal component of output in inverse fourier form   $$ y'(t) = \int_{-\infty}^{\infty}  S(f) H(f)  \ e^{j2pit}\ df $$ where $S(f)$ and $H(f)$ are spectrums of input signal $s(t)$ and Frequency response of filter. 

Output SNR of filter which is sampled at time instant $\,t=T $ is output power with signal component by noise power which can be return as 

$$SNR  =  \frac{\displaystyle\left |\int_{-\infty}^{\infty}  S(f) H(f)  \ e^{j2piT}\ df \right|^2}{ \displaystyle\left|\int_{-\infty}^{\infty}  \frac{N_o}{2} |H(f)|^2  df \right|^2}$$

Note that numerator term is evaluated at time t = T, which is the sampling instant for Matched filter. 

To maximise this SNR , we need to cancel the phase term . 
if we substitute $$H(f) = S^*(f)  e^{-j2pif} $$
which is nothing but $$h(t) = s^*(T-t) $$

and simplify the above SNR expression, we will get 
$$ SNR = \frac{\displaystyle\int_{-\infty}^{\infty} \left|S(f)\right|^2 df}{N_o/2} $$
which is nothing but ration of energy of signal $s(t)$ to value of power spectral density of AWGN noise. 

In this time domain analysis you can see that there is no filter effect , and it also tells us what should be the filter impulse response to maximise the output SNR.