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8 proves the conclusion. 1. Let µ be a Radon measure on Rn . 5) holds, then x is called a Lebesgue point. In fact, for each rational number cj , we can find a set Ej such that µ(Ej ) = 0 and lim r→0 1 µ(B(x, r)) B(x,r) |f (y) − cj | dµ(y) = |f (x) − cj | 44 Measure theory for every x ∈ Rn − Ej . 5) holds for every j x ∈ Rn − E. 6. If f is integrable on Rn , then |f (x + h) − f (x)| dx = 0. lim h→0 Proof. 3 we take ϕ ∈ C0 (Rn ) for which |f (x) − ϕ(x)| dx < ε. Clearly, |ϕ(x + h) − ϕ(x)| dx = 0. lim h→0 Hence we have |f (x + h) − f (x)| dx |f (x + h) − ϕ(x + h)| dx |ϕ(x + h) − ϕ(x)| dx + + |f (x) − ϕ(x)| dx + = 2 |ϕ(x) − f (x)| dx |ϕ(x + h) − ϕ(x)| dx, which gives lim sup h→0 |f (x + h) − f (x)| dx < 2ε.

Consider the family U of all nonnegative functions u ∈ Lp (µ) such that λ+ (f ) whenever f ∈ Lp (µ) and f f u dµ 0. Then we see that µ(p ) (u) A, if we consider f = (u/µ(p ) (u))p −1 . If u and v are in U, then we claim that u ∨ v = max{u, v} are also in U. In fact, considering the characteristic function g on the set {x : u(x) v(x)}, we have for any nonnegative p function f ∈ L (µ), λ+ (f ) = λ+ (f g) + λ+ (f (1 − g)) f gu dµ + (f (1 − g))v dµ = f (u ∨ v) dµ. Now we find an increasing sequence {uj } of functions in U for which lim µ(p ) (u) = sup {µp (u) : u ∈ U}.

Let µ be a nonnegative Radon measure, and 1 < p < ∞. If {fj } is bounded in Lp (µ), then there exists a subsequence {fjk } which converges weakly in Lp (µ). 2, we can find a countable dense subset {gk } of Lp (µ). Since {g1 (fj )} is bounded, find a subsequence {f1,j } for which {g1 (f1,j )} converges. Next, find {f2,j } {f1,j } for which {g2 (f2,j )} converges. Repeating this process, we obtain {f1,j } {f2,j } ··· {fk,j } ··· for which {gk (fk,j )} converges as j → ∞. 2. 2. Another application of the method of diagonal sequence is the following result, which is known as the Ascoli-Arzel`a theorem.