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If two signed measures µ, ν can be supported on disjoint sets, we say that they are mutually singular (or that µ is singular with respect to ν) and write µ ⊥ ν. 2. 5 (Jordan decomposition theorem). Every signed measure µ an be uniquely decomposed as µ = µ+ − µ− , where µ+ , µ− are mutually singular unsigned measures. ) We refer to µ+ , µ− as the positive and negative parts (or positive and negative variations) of µ. This is of course analogous to the decomposition f = f+ − f− of a function into positive and negative parts.

Real analysis (ii) (Homogeneity) cf c. Lp = |c| f Lp for all complex numbers (iii) ((Quasi-)triangle inequality) We have f +g Lp ≤ C( f Lp + g Lp ) for some constant C depending on p. If p ≥ 1, then we can take C = 1 (this fact is also known as Minkowski’s inequality). Proof. The claims (i), (ii) are obvious. 16) and is left as an exercise. 16). By the non-degeneracy property we may take f Lp and g Lp to be non-zero. Using the homogeneity, we can normalise f Lp + g Lp to equal 1, thus (by homogeneity again) we can write f = (1 − θ)F and g = θG for some 0 < θ < 1 and F, G ∈ Lp with F Lp = G Lp = 1.

We will not take this approach here, but see for instance [LiLo2000] for a discussion. 15. 14 below) in the special case when f , g are generalised step functions, say f = A1E and g = B1F with A, B non-zero. 24) µ(E ∩ F )1/r ≤ µ(E)1/p µ(F )1/q which can be easily deduced from the hypothesis p1 + 1q = 1r and the trivial inequalities µ(E ∩ F ) ≤ µ(E) and µ(E ∩ F ) ≤ µ(F ). 24) only holds if µ(E ∩ F ) = µ(E) = µ(F ), or in other words if E and F agree almost everywhere. Note the above computations also explain why the condition p1 + 1q = 1r is necessary.

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