If we assume that m(t) = c(t) = 0 for t < 0, we can show analytically that the convolution integral m(t)*c(t) is periodic with period 1/fc, and one period is
p(t) = (sin(2*pi*fs*t)/(2*pi*fs) , t < (1/(2*fc)
p(t) = 0, 1/(2*fc) < t < 1/fc
When approximating the convlution integral with the convolution sum, don't forget to scale the sum by dt. Here is the code, extending the time vector furher out
fs=10^6;
T=1/fs;
tt=0:T/100:60*T;
m=cos(2*pi*fs*tt);
fc=10^5;
c=0.5*(1+square(2*pi*fc*tt));
yy=conv(m,c)*tt(2);
pfunc = @(t) (sin(2*pi*fs*mod(t,1/fc))/(2*pi*fs).*(mod(t,1/fc) < 1/2/fc));
plot(tt,yy(1:numel(tt)),tt,pfunc(tt))
As you can see, the approximating convolution sum isn't quite accurate and becomes less so as t increases. I'm not exactly sure why that is, some sort of round-off accumulation perhaps?
