Why $\|f-g\|=0$ if and only if $f=g$?

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I'm learning Fourier Transformation lately, and in the Course Reader page 23, it defines $\|f\|=\left(\int_0^1 \left|f(t)\right|^2 dt\right)^{1/2}$.

And then $\|f-g\|=0$ if and only if $f=g$.

My question is what does it mean exactly by $f=g$? Is it a new definition of $f=g$? Since, for example $f=0$ and $g(x)=1$ for $x=0.5$ and $0$ otherwise, will result $f=g$.