functional analysis: show L^1 integral operator has norm 1

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I just started my course in functional analysis and have already stumbled across some things I don't understand, which are quite basic :(. In my lecture notes it says: Let $\mu$ be a measure on a measurable space $(\Omega, \mathscr F)$. We define the operator $Q: L^1(\mu) \to \Bbb R$, by $Q(f)=\int_{\Omega} f d\mu$, for $f \in L^1(\mu)$. It follows that the operator $Q$ is linear and boundend and has norm 1.

I have already verified linear and bounded. However, I get stuck on showing that it has norm 1. Here is what I have done:

$\lVert Q(f)\rVert= sup_{f \in L^1(\mu)}\lvert \int_{\Omega} f d\mu\rvert \le sup_{f \in L^1(\mu)}\int_{\Omega}\lvert f \rvert d\mu = \lVert f \rVert_{L^1(\mu)}$, thus $\lVert Q(f)\rVert \le 1$.

Now, I do not really understand how to tackle the reverse inequality. I was hoping perhaps someone could help me understand this basic notion.

Thanks a lot and wish you a good day.

Frederique