The Lebesgue Integral

Slice the range, not the domain, and integrate what Riemann cannot.

The idea

The Lebesgue integral measures the region under a function by cutting the range into horizontal bands, where the Riemann integral cuts the domain into vertical strips: for each value level, it asks how much of the domain the function carries to that level. Write $\mu(A)$ for the measure of a set $A$ — for an interval, just its length — and assume throughout that the sets and functions in play are measurable, so that $\mu$ applies to them.

The integral is built in three stages. A simple function takes only finitely many values, each on its own measurable set: $\varphi = \sum_{i} c_{i}\,\mathbf{1}_{A_{i}}$ with the $A_{i}$ disjoint. Its integral is defined, with no limit involved, as each value times the measure of the set carrying it, $\int \varphi\,d\mu = \sum_{i} c_{i}\,\mu(A_{i}).$ For a non-negative measurable $f$, take the supremum over the simple functions below it, $\int f\,d\mu = \sup\left\{\, \int \varphi\,d\mu \;:\; \varphi \text{ simple},\ 0 \le \varphi \le f \,\right\},$ a value in $[0, \infty]$. For a general measurable $f$, split it into $f^{+} = \max(f, 0)$ and $f^{-} = \max(-f, 0)$, both non-negative, and set $\int f\,d\mu = \int f^{+}\,d\mu - \int f^{-}\,d\mu$. The subtraction is defined exactly when both pieces are finite, which is the condition, $\int |f|\,d\mu < \infty$, for calling $f$ integrable.

Wherever a function is Riemann integrable, it is Lebesgue integrable with the same value, so the new integral extends the old. And a set of measure zero contributes nothing to the sums above, so changing a function on such a set changes none of its integrals.

Ways to work on it

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