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Exercise 1.9 (Basic properties of the complex integral)

Let f,g : Ω → C be real-valued absolutely integrable functions, and let c ∈ C.

1.
(Linearity) Show that f + g and cf are also complex-valued absolutely integrable functions, with ∫ Ωf + gdμ = ∫ Ωfdμ + ∫ Ωgdμ

and

∫ Ωcfdμ = c∫ Ωfdμ.

(For the second relation, one may wish to first treat the special cases c > 0 and c = −1.)

2.
(Equivalence) Show that if f and g are equal almost everywhere, then ∫ Ωfdμ = ∫ Ωgdμ.

3.
Show that ∫ Ω|f|dμ ≥ 0, with equality if and only if f is zero almost everywhere.
4.
(Markov inequality) Show that μ({ω : |f(ω)|≥ t}) ≤ 1 t ∫ Ω|f|dμ for any 0 < t < ∞.

Answers

1.
(Linearity)
Since the complex simple integral can be written in terms of two real valued functions (f + g) = ℜ(f + g) + i ⋅ℑ(f + g) = [ℜ(f) + ℜ(g) ] + i ⋅ [ℑ(f) + ℑ(g) ]

Notice that both ℜ(f) + ℜ(g) and ℑ(f) + ℑ(g) are real valued functions; applying the previous part of the proof we get the desired result.
The scalar multiplication property is demonstrated as follows. We can rewrite cf as

cf(x) = [ℜ(c)+iℑ(c)]⋅[ℜ(f(x))+iℑ(f(x))] = ℜ(c)⋅ℜ(f(x))−ℑ(c)ℑ(f)+iℑ(c)ℜ(f)+iℜ(c)ℑ(f)

Using this identity we obtain

cf = ℜ(cf) + i ⋅ℑ(cf) = [ℜ(c)ℜ(f) −ℑ(c)ℑ(f) ] + i ⋅ [ℑ(c)ℜ(f) + ℜ(c)ℑ(f) ] = ℜ(c)ℜ(f) −ℑ(c)ℑ(f) + iℑ(c) ⋅ℜ(f) + ℜ(c)ℑ(f) = ℜ(f) ⋅ [ℜ(c) + iℑ(c)] + iℑ(f) [ℜ(c) − 1∕iℑ(c)] = [ℜ(c) + iℑ(c)] ⋅ (ℜ(f) + iℑ(f)) = c ⋅ f
2.
(Equivalence)
Equivalence follows by applying the analogous property of the real valued functions separately to the real and imaginary part of the complex simple integral.
3.
Since |f| = ℜ(f)2 + ℑ(f)2 is an unsigned function, the assertion follows by the previous exercise.
4.
Just as in the real-valued case, consider the trivial pointwise inequality λ1{x∈Ω:f(x)≥λ}≤ f(x).

We apply integral from both sides and using the compatibility of Lebesgue measure with the Lebesgue integral (Exercise 1.5), we conclude

λm({x ∈Ω : f(x) ≥ λ}) ≤ f.

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2021-08-15 00:00
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