Answer: The Lebesgue Integral That Measures Its Own Level Sets

Key Idea / Intuition

Instead of slicing the domain horizontally (the usual Riemann picture), slice it vertically in the range: the integral of a non-negative function equals the "area under its graph," and that area can be computed by stacking horizontal slices of width $dt$, each slice having length equal to the measure of the level set $\{f > t\}$. This is the layer-cake (Cavalieri) representation — a fundamental alternative way to think about integration that replaces knowing $f$ pointwise with knowing the sizes of its superlevel sets.

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Formal Proof / Solution

Step 1: Prove the layer-cake formula

Consider the product space $[0,1] \times [0,\infty)$ with the product measure $m \times m$. Look at the region under the graph:

$E = \{(x, t) : x \in [0,1],\; 0 \leq t < f(x)\}.$

Since $f$ is measurable, $E$ is a measurable subset of $[0,1] \times [0,\infty)$.

Compute $m \times m(E)$ by slicing in $x$:

For each fixed $x$, the slice $\{t : 0 \leq t < f(x)\}$ has measure $f(x)$. So by Fubini/Tonelli:

$m \times m(E) = \int_0^1 f(x)\, dx.$

Compute $m \times m(E)$ by slicing in $t$:

For each fixed $t \geq 0$, the slice $\{x \in [0,1] : t < f(x)\} = \{f > t\}$ has measure $m(\{f > t\})$. So:

$m \times m(E) = \int_0^\infty m(\{x : f(x) > t\})\, dt.$

Since both expressions equal $m \times m(E)$, we conclude:

$\boxed{\int_0^1 f(x)\, dx = \int_0^\infty m(\{f > t\})\, dt.}$

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Step 2: Apply it to $f(x) = x^n$

We need to compute $m(\{x \in [0,1] : x^n > t\})$ for each $t \geq 0$.

• If $t \geq 1$: the set $\{x^n > t\}$ is empty (since $x \leq 1$), so measure $= 0$.
• If $0 \leq t < 1$: the condition $x^n > t$ means $x > t^{1/n}$, so the set is $(t^{1/n}, 1]$, which has measure $1 - t^{1/n}$.

Therefore:

$\int_0^\infty m(\{x^n > t\})\, dt = \int_0^1 \left(1 - t^{1/n}\right) dt$

$= \left[t - \frac{t^{1/n + 1}}{1/n + 1}\right]_0^1 = 1 - \frac{1}{1/n + 1} = 1 - \frac{n}{n+1} = \frac{1}{n+1}.$

So we recover $\displaystyle\int_0^1 x^n\, dx = \frac{1}{n+1}$ without ever computing an antiderivative of $x^n$ directly — just by measuring level sets!

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Why this is beautiful

The layer-cake formula is not just a trick — it is the conceptual foundation for:

• The definition of the Lebesgue integral via the distribution function,
• Lp interpolation and norm identities,
• Geometric inequalities like the Brunn–Minkowski theorem.

It says: to integrate $f$, you don't need to know $f$ itself — knowing the size of its superlevel sets (its distribution) is enough.