The top of the Potential Divider is 5V; the bottom is at 0V and the middle (connected to A0) is some value between 5V and 0V that varies as the LDR resistance varies according to the light level.

As both resistors are in series the same current must flow through them both. The LDR resistance drops with light, which causes the current in both resistors to increase (I=V/R), and therefore the voltage across the other (non-LDR) increases.

So as the LDR resistance varies with Light, the Voltage at A0 will too.

Technically you can work out the Voltage on A0 as :

Va0 = 5 * R1/(R1+R2)

where Va0 is the voltage at A0 pin, R2 is the top resistor value, R1 is the bottom resistor value;

e.g. R1 = 10k, R2 = 5k => Va0 = 5 * 10000/(10000 + 5000) = 5 * 10/15 = 3.33V

The LDR has a high value when no light is present. The value of resistance of the LDR depends on the type. In this case it's about 10k. As the light level increases the resistance drops, which makes the current increase (by Ohm's Law), which in turn, makes the voltage at A0 (Va0) increase.

Electrically this is what's happening (skip this if you're not interested in this): The LDR and resistor are in series with the applied voltage (5V), so the current flowing through them is the same (the A0 pin draws virtually zero current). So the current through the resistor is (by Ohm's Law):

I = 5 / (R1+R2)

Now, the voltage across the resistor is applied to A0. Again, by Ohm's Law that is:

Va0 = I * R1

Substituting the equation for I back in to this equation we get:

Va0 = 5 * R1/(R1+R2)

See what it looks like on a Breadboard.