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The higher resistance of the multimeter fuses is probably because they are super fast acting (FF), therefore they need to dissipate more power to melt in a shorter time.
Yes, power dissipation is a big part of the regular operation of the fuses. They need that power to burn through. FF plays a role in here, but also the HRC voltage. The HRC versions of fuses use sand to extinguish arcing. That also acts as a heat sink by design.
These two are reasons why they need larger voltage (power) than a cheap glass fuse
Another thing that can bite you is the thermo electric effect if you are not careful with the PCB layout and one end of the shunt resistor has a different temperature than the other end. When dealing with signals in the uV range this is a serious thing to consider.
Meters are essentially a sealed unit and the internals are pretty close to thermal equilibrium, so not an issue in practice.
But yes, the lower you go, the more this is an issue. A big issues in the likes of the 6.5/7.5 digit high end meters.
When you run the ampmeter at the top of the protected range, the fuse is at it's maximum power dissipation. Short before burnout.
You don't want that anywhere near the sense wires of a microvolt measurement. . The fuse is also a major violation to the thermal symmetry (equilibrium) of your circuit.
In general, the current measurement with a multimeter (CAT-) rated for mains voltages or even higher seems like a (bad) compromise to me.
As a more pragmatic solution to the original problem (current measurement for a low voltage but high current device), I would rather like to have a power supply with external sense wires. This should be much cheaper to achieve than a SMU.
Measuring milli- or microvolts across an external current shunt (right in the circuit under test) is probably a more precise and clear solution than gilding the Lilly inside the multimeter.