ESR is a series resistance in series with the capacitor. Put in mind that you would place a 1 ohm resistor in series with a capacitor to decouple your power supply. Would you think that would improve the effect of that capacitor? There is a reason to avoid capacitors with such high ESR values.

A capacitor is designed to block DC (high impedance) and short high frequencies (low impedance). Impedance is a complex number that consists of resistance and reactance. It is written as Z=R + jX. Here is R the 'real part' and X the 'imaginary part' (note the j symbol).

A resistor has an impedance of Z=R. A capacitor has an impedance of Z=1/jwC. As you can see, a capacitor is has an imaginary component. This is for a perfect capacitor. Here is C the capacitance value, and w the angular velocity (=frequency times 2*PI).

But we are measuring imperfect capacitors (that's what an ESR meter is for), so the impedance will be Z = R_ESR + 1/jwC. R_ESR is the resistor in series with the capacitor. If you want to know the 'resistance' (better to say: impedance) of the component at a certain frequency (capacitors are frequency dependant), you need to calculate the magnitude:

|Z| = sqrt(R^2 + X^2)

So, for a capacitor that is:

|Z| = sqrt(R_ESR^2 + (1/wC)^2)

If you apply a sine wave of 1Vpp on a capacitor and measure the current at different frequencies, you should get different values of current. That is because the impedance of the capacitor is moving from very high (blocking DC currents) to very low (shorting high frequency AC).

->> Now after this boring theory, it comes down to this: If you apply an AC voltage and you make w (the frequency) very high, it will mean that the piece 1/wC of the impedance becomes very small. If you divide 1 by a big number, you get a small number. If you raise this frequency so much that the part 1/wC is ignorable, you can measure the ESR value (because the impedance of what you're measuring only shows the ESR). After all, if you have something like:

|Z| = sqrt(0.1^2 + 1/(2*PI*150k*4700u)^2)

|Z| = sqrt(0.01 + 5.096383*10^-8)

with R ESR = 0.1 ohms, C = 4700uF, test frequency of 150kHz. The impedance is:

= 0.1000000254 (ohms, for 150kHz signals).

If you want to see the effect of the capacitor at 150kHz (and C=4700uF), you need atleast a 8-digit multimeter. Goodluck

If you would do this at like say, 500Hz:

|Z| = sqrt(0.1^2 + 1/(2*PI*500*4700u)^2)

|Z| = sqrt(0.01 + 0.00458674)

=0.120776 (ohms, for 500Hz).

Which is a value we could expect, because a capacitor tries to block low frequent signals.