# Is null function equivalent to null integral?

If $f:X\to\mathbb R$ is integrable in the measure space $(X,\Sigma,\mu)$, are these equivalent?

1. $f\equiv0$ $\mu$-almost everywhere.
2. $\displaystyle\int_E \;\,f \;\,\mathrm d\mu=0$ for all $E\in\Sigma$.

$(1)\Rightarrow(2)$ seems obvious. Here's my (possibly hasty and flawed) rationale for $(2)\Rightarrow(1)$.

Let $g\ge0$. So $\int gI_E \;\mathrm d\mu=0$ if $E$ is measurable. Since $I_E$ is also non-negative, and $E$ is arbitrary, $gI_E\equiv0$ for all $E$, which implies $g\equiv0$ a. e.

For arbitrary integrable $f$, define $f_+\triangleq \max\{0,f\}$ and $f_-\triangleq \max\{0,-f\}$. Then $\int_E\; f\;\mathrm d\mu =\int_E\; f_+\;\mathrm d\mu -\int_E\; f_-\;\mathrm d\mu =0$. $E$ is arbitrary, therefore, $\int_E\; f_+\;\mathrm d\mu =\int_E\; f_-\;\mathrm d\mu =0$, and because these functions are non-negative, the previous paragraph indicates that $f_+\equiv f_-\equiv 0$, and finally $f\equiv0$ $\mu$-almost everywhere.

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How did you get $\int_E\; f_+\;\mathrm d\mu =\int_E\; f_-\;\mathrm d\mu =0$ in the last paragraph? – Ashok Nov 12 '11 at 8:11

Let $E^+=\{ x : f(x)>0\}$ and $E^-=\{x: f(x) <0\}$. Then, assuming (2) holds, $\int_{E^+} f =0=\int_{E^-} f$. This implies that $f_+$ and $f_-$, as defined in your post, are both 0-valued a.e. It then follows that $f$ is 0-valued a.e.