# $f_1,f_2 \in L^q(\mu)$ and $\int_\mathcal{X}f_1gd\mu = \int_\mathcal{X}f_2gd\mu$ for all $g \in L^p(\mu)$ implies $f_1=f_2$ a.e.

Let $X=(\mathcal{X},\mathcal{M},\mu)$ be a measure space. Assume that $\mu$ is $\sigma$ finite and $1\leq p \leq \infty$, with $q$ the Holder conjugate exponent. If $f_1,f_2 \in L^q(\mu)$ and $\int_\mathcal{X}f_1gd\mu = \int_\mathcal{X}f_2gd\mu$ for all $g \in L^p(\mu)$ show that $f_1=f_2$ a.e.

I think I can use Holder's inequality to show that $f_1g$ and $f_2g$ are in $L^1(\mu)$. Then if I could prove that $\int_\mathcal{X}|(f_1-f_2)g|d \mu=0$ then since is must hold for all $g$ then $f_1=f_2$ a.e. However I can only get $|\int_\mathcal{X}(f_1-f_2)gd \mu|=0$. Probably I need to use the $\sigma$ finite statement somewhere but I can't find how. Can anyone give me any hints? Is my approach correct? Thank you very much guys!

• Try $g = sign((f_1-f_2) \cdot |f_1-f_2|^{1/(p-1)}$. – Hans Engler Apr 20 '15 at 2:13
• I don't think I understand what you trying to say. – SpawnKilleR Apr 20 '15 at 2:15
• Just use this $g$ and work out $\int_X (f_1-f_2)g$. What is the result? – Hans Engler Apr 20 '15 at 2:17
• Sorry - this was supposed to be $g = sgn(f_1-f_2)|f_1-f_2|^{p-1}$. Check whether this $g$ is in $L^q$. – Hans Engler Apr 20 '15 at 2:21

It is enough to prove that if $f\in L^q(\mu)$ is such that $\int_{\cal X}fg=0$ for all $g\in L^p(\mu)$, then $f=0$ a.e.
Let $g_1:{\cal X}\to \mathbb{R}$ be defined by $$g_1(x)=\begin{cases} f(x)|f(x)|^{q-2} &\mbox{ if } f(x)\ne 0\\ 0 &\mbox{ otherwise } \end{cases}$$ We have $$\int_{\cal X}|g_1|^p\,d\mu=\int_{\{f\ne 0\}}|f|^{p(q-1)}\,d\mu=\int_{\{f\ne 0\}}|f|^q\,d\mu=\int_{\cal X}|f|^q\,d\mu,$$ i.e. $g_1\in L^p(\mu)$.
Now we have $$\int_{\cal X}fg_1\,d\mu=\int_{\cal X}|f|^q=0 \Rightarrow f=0 \,\mbox{ a.e.}.$$ Replacing $f$ by $f_1-f_2$ we get the desired result.