State and prove Lusins theorem

Answer 1

Lusin's theorem says that every measurable function f is a continuous function on nearly all its domain.

It is given that f measurable. This tells us that it is bounded on the complement of some open set of arbitrarily small measure. Now we redefine ƒ to be 0 on this open set. If needed we can assume that ƒ is bounded and therefore integrable.

Now continuous functions are dense in L1([a, b]) so there exists a sequence of continuous functions an tending to ƒ in the L1 norm. If we need to, we can consider a subsequence.
We also assume that an tends to ƒ almost everywhere. Now Egorov's theorem tells us that that an tends to ƒ uniformly except for some open set of arbitrarily small measure. Since uniform limits of continuous functions are continuous, the theorem is proved.

Answer 2

Lami's theorem states that if three concurrent forces act on a body keeping it in Equilibrium, then each force is proportional to the sine of the angle between the other two forces.

Let P, Q and R be the three forces acting at a point O.

Since OP, OQ and OR are vectors, they can be arranged "nose to tail". Then since the forces are in equilibrium, they form a closed triangle.

Applying the sine rule to the triangle, P/sin[pi - angle(QOR)] = Q/sin[pi- angle(ROP)] = R/sin[pi- angle(POQ)]

Then, since sin(pi - x) = sin(x)

P/sin[angle(QOR)] = Q/sin[angle(ROP)] = R/sin[angle(POQ)]