Meaning of equality in $L^p$ Let $(\Omega, \mathcal A, \mu)$ be a measure space. What does it mean for two functions $f,g$ to be equal in $L^p(\Omega, \mathcal A, \mu)$ ? Does that mean $\|f-g\|_{L^p}=0$ ? Or that $f-g=0 $ $\mu$-a.e ? If the later is the right answer, what is the difference between being equal in $L^p$ and $L^{p+1}$ ? I mean shouldn't the definition depend on $p$ ? 
 A: If $f,g \in L^p$ then $f = g$ in $L^p$ if and only if $f = g$ a.e. which is if and only if $\|f-g\|_{L^p} = 0$. 
The definition of being equal in $L^p$ depends on $p$ only in the sense that you require that the functions themselves are in $L^p$ to say that they are equal in $L^p$.  
A: For a measurable function $f:\Omega\to\mathbb{R}$, the condition $\|f\|_{L^p}=0$ does not imply that $f\equiv0$. It only implies that $f=0$ almost everywhere. This means that $\|\cdot\|_{L^p}$ cannot be used as a norm for measurable functions. However, it also points to the solution: We need to identify two functions if they are equal to each other almost everywhere, and consider $L^p(\Omega, \mathcal A, \mu)$ as the set consisting of those equivalence classes.
A: The difference between $L^p$ and $L^q$ with $p>q$ is not about functions being equal or not, but in how much two functions are "near" to each other and in whether some sequences of functions converge to any. For example, the sequence $f_n(x)=\frac{1}{n}\cdot 1_{[0,n]}(x)$ converges to zero in $L^2$, but is not a Cauchy sequence in $L^1$ (and therefore cannot converge).
