When doing ACSR |=(1<<ACD)|(1<<ACBG) are the rest of the bits assumed to be 0?

[johnwasser]

@jerive

ACSR |= (1<<ACD) | (1<<ACBG);

You are setting those two bits and leaving the others unchanged. If you want to set those two and clear the rest:
ACSR = (1<<ACD) | (1<<ACBG);

Note: In the Analog Compare Status Register, Bit 5 (ACO) is read-only. It doesn't matter what you try to write to it.

Note: In the Analog Compare Status Register, Bit 4 (ACI) is an interrupt flag. It gets CLEARED if you write a 1 to it.