If $f\in \mathcal{L}_+^0$ and $\int fd\mu=0$, where $\mathcal{L}_+^0$ is the set of all measurable $[0,\infty]$-valued functions, show that $f=0$ almost everywhere.

Is this correct?

$$\cap_n\left\{x\mid|f|≥n^{−1}\right\}⊆\frac{1}{n}\left\{x\mid|f|≥n^{−1}\right\}⟹\mu (\cup_n\left\{x\mid|f|≥n^{−1}\right\})\leq \mu(\frac{1}{n}\left\{x\mid|f|≥n^{−1}\right\})\leq \int_{|f|≥1/n}f dμ=0$$

If not please direct me in the correct approach.

  • $\begingroup$ Assume it is non-zero on a set of non-zero measure, and try to derive a contradiction. $\endgroup$ Oct 7 '16 at 12:59

Define $E\subset \mathbb{R}$ to be the set of points where $f$ is non-zero and assume towards a contradiction that $\mu(E)>0$.

Then we have $$\int_{\mathbb{R}}f d\mu = \int_{(\mathbb{R}-E)\cup E}fd\mu = \int_{(\mathbb{R}-E)} fd\mu + \int_{E}f d\mu = \int_{E}f d\mu$$

Since we defined the set $E$ to be points where $f$ is non-zero, and since $f>0$ by construction, it must be the case that $$\int_E fd\mu >\int_E 0 d\mu = 0$$

Putting this together, we have shown that $$\int_{\mathbb{R}}f d\mu >0$$

which is a contradiction. This must mean that $\mu(E) = 0$, which means that $f=0$ a.e.


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