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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.
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)]
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