Positive and negative parts
So far the integral is defined only for non-negative functions. To handle a function f that goes both positive and negative, split it. Define f^+ = max(f, 0) and f^- = max(-f, 0). Both are non-negative and measurable, and they reconstruct f cleanly: f = f^+ - f^- and |f| = f^+ + f^-.
f^+(x) = max(f(x), 0) f^-(x) = max(-f(x), 0)
Both >= 0, both measurable. Then for every x:
f(x) = f^+(x) - f^-(x)
|f(x)|= f^+(x) + f^-(x)
Example. f(x) = sin x on [0, 2pi]:
f^+ = sin x on [0, pi], 0 on [pi, 2pi]
f^- = 0 on [0, pi], -sin x on [pi, 2pi]
integral f^+ = 2, integral f^- = 2
integral f = integral f^+ - integral f^- = 2 - 2 = 0.The definition of integrable, and L1
We say f is [[integrable-function|integrable]] when both integral f^+ and integral f^- are finite, and then set integral f = integral f^+ - integral f^-. The single clean test for this is: f is integrable exactly when the integral of |f| is finite. The collection of all such f is the L1 space — the summable functions. To subtract the two pieces we must avoid the meaningless +infinity minus +infinity; demanding integral |f| < infinity is precisely what rules that out.
Linearity, and ignoring null sets
On L1 the integral is at last linear: integral(af + bg) = a·integral f + b·integral g for real a,b. The linearity of the integral is proved by reducing to the non-negative case and rearranging positive and negative parts (using the additivity that monotone convergence guarantees). And integration respects equality up to null sets: if f = g almost everywhere — that is, except on a set of measure zero — then integral f = integral g. Sets of measure zero are simply invisible to the integral.
A useful corollary, the triangle inequality for integrals: |integral f| <= integral |f|, valid for every f in L1. It follows by writing integral f = integral f^+ - integral f^- and comparing with integral f^+ + integral f^- = integral |f|. We will lean on this bound throughout the convergence theory.