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**Unformatted text preview: **≥ 0, we know that E(X + ) ≥ 0 and E(X − ) ≥ 0. We now use the fact that if
the sum of two non-negative numbers is ﬁnite, then each number must be ﬁnite to conclude
that E(X + ) < ∞ and E(X − ) < ∞. Thus, by deﬁnition, E(X ) = E(X + ) − E(X − ) < ∞ so
that X ∈ L1 .
Corollary. If X ∈ L1 , then |E(X )| ≤ E(|X |). In particular, if E(|X |) = 0, then E(X ) = 0.
Proof. Since E(X + ) ≥ 0 and E(X − ) ≥ 0, we have from the triangle inequality
|E(X )| = |E(X + −X − )| = |E(X + )−E(X − )| ≤ |E(X + )|+|E(X − )| = E(X + )+E(X − ) = E(|X |).
Since |E(X )| ≥ 0, if E(|X |) = 0, then 0 ≤ |E(X )| ≤ E(|X |) = 0 implying E(X ) = 0.
Theorem 20.3. If X, Y : (Ω, F , P) → (R, B ) are integrable random variables with X = Y
almost surely, then E(X ) = E(Y ).
Proof. Suppose that X = Y almost surely so that P {ω ∈ Ω : X (ω ) = Y (ω )} = 1. To begin,
assume that X ≥ 0 and Y ≥ 0 and let A = {ω : X (ω ) = Y (ω )} so that P {A} = 0. Write
E ( Y...

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