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Prove that if the integral of nonnegative function is zero then the function is zero

Let f be an integrable, non-negative function on the interval [a,b]. Prove that if

    \[ \int_a^b f(x) \, dx = 0 \]

then f(x) = 0 at each point x \in [a,b] at which f is continuous.


Proof. The proof is by contradiction. Let p \in [a,b] be a point at which f is continuous, and suppose f(p) \neq 0. This implies f(p) > 0 since f is non-negative by hypothesis. Since f is continuous at p, we know by the sign-preserving property of continuous functions (Theorem 3.7 in Apostol), that there is some neighborhood about p, say (p - \delta, p + \delta) such that f(x) has the same sign as f(p) for all x \in (p - \delta, p + \delta), i.e., such that f(x) > 0 for all x \in (p - \delta, p + \delta). But then by the monotone property of the integral we know,

    \[ \int_{p - \delta}^{p + \delta} f(x) \, dx > \int_{p - \delta}^{p + \delta} 0 \,dx = 0.\]

But then,

    \[ \int_a^b f(x) \,dx = \int_a^{p - \delta} f(x)\, dx + \int_{p - \delta}^{p + \delta} f(x) \, dx + \int_{p + \delta}^b f(x) \, dx = 0. \]

This is a contradiction since f is nonnegative we know

    \[ \int_a^{p - \delta} f(x) \, dx \geq 0 \qquad \text{and} \qquad \int_{p + \delta}^b f(x) \, dx \geq 0. \]

Since \int_{p - \delta}^{p + \delta} f(x) \, dx is strictly positive, the sum of the three pieces of the integral \int_a^b f(x) \, dx cannot equal 0. \qquad \blacksquare

One comment

  1. Jindrich Michalik says:

    The theorem and the proof are good and useful, but I suggest adjusting the title. The note about continuity is important.

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