If $\int f=0$, can I conclude that $f=0$?

Question

Suppose, $f:(0,1)\to\Re_+$ and $f\in L((0,1))$, therefore measurable. Is the following implication correct? $$\int_{(0,1)}f(x)\,d\mu=0\overset{?}{\implies} f(x)=0$$

Intuitively, I feel like the conclusion should hold because $f$ is a non negative function, but is there anyway of showing this?

No, we can only say that $f = 0$ almost everywhere, although if $f$ is also continuous, then the implication does hold, in the sense that $f = 0$. — Jakobian, Nov 19 '19 at 10:47

marwalix · Accepted Answer · 2019-11-19T11:04:35.343

2

Take

$$f(x)=\begin{cases} &{1\over q}\quad x={p\over q}\in\Bbb{Q}\\ &0\quad x\in\Bbb{R}-\Bbb{Q} \end{cases}$$

This function is Lebesgue integrable with integral $0$

edited Nov 19 '19 at 11:04

answered Nov 19 '19 at 10:50

marwalix

16,773

A few notes about formatting: when writing cases, you don't need an equals sign for each case. You can just do $f(x)=\begin{cases}1/q\0\end{cases}$, which simplifies things. Also, if you want a space of a certain size in mathjax, there's different codes for each space. e.g. ,,,,,, = \quad. – Jam Nov 19 '19 at 10:56
Thanks for the formatting tips – marwalix Nov 19 '19 at 11:03
@marwalix this is somewhat related, if $f(x)>0$ can I conclude $\int f(x)>0$? – DMH16 Nov 19 '19 at 11:10
Yes you can. Tedious to prove in the general setup – marwalix Nov 19 '19 at 11:16

If $\int f=0$, can I conclude that $f=0$?

1 Answers1