Is $(f\mapsto\int_\Omega f d\mu)\in L^1(\Omega,\lambda)^*$? For $\Omega\subset\mathbb{R}^n$ compact and some regular Borel measure $\mu\colon\mathfrak{B}(\Omega)\to\mathbb{R}$, is the mapping $$f\mapsto\int_\Omega f d\mu$$ a linear and continuous functional on $L^1(\Omega,\mathfrak{B}(\Omega),\lambda)$?
Can we show that the integral expression is bounded, or can we find a counterexample to it?
Edit: To avoid misunderstandings I would like to emphasize again that we are dealing here with two different measures $\lambda$ and $\mu$. Primarily I am looking for a counterexample of an $L^1$-function $f$ (with respect to $\lambda$) and a measure $\mu$ such that the integral of $f$ w.r.t. $\mu$ is no longer finite.
I have already played around with various $L^1$ functions and measures, but have never come to a contradiction.
 A: You don't need to decide whether the integral of $f$ is finite or not. If $\int_\Omega f~\mathrm d\mu$ is not finite, then $f$ is not $L^1$ ! Now that we know all these integrals are finite, the linearity of the integral shows this map is linear, and hence is a member of the dual space of $L^1(\Omega,\mathfrak{B}(\Omega),\mu)$.
A: Consider any point $x_0 \in \Omega$ and let $\delta_{x_0}$ be the Dirac measure, i.e. $\delta_{x_0}(A) = 1(x_0 \in A)$ for any $A \in \mathfrak{B}(\Omega)$. This is a Borel regular measure. Define the function $f(x) = 1(x = x_0)$, so $f(x) = 1$ for $x = x_0$ and $f(x) = 0$ otherwise. Then
$$
\int f d\delta_{x_0} = 1
$$
but
$$
\int |f| d\lambda = 0.
$$
A: It is a basic fact that integration $I : L^1(\Omega, \mu) \to \mathbb{C}$ is a linear functional. It remains to bound the absolute value of the integral to prove that $I$ is continuous. This is straightforward since we have the identity
$$\left|\int_{\Omega}f\,d\mu\right| \leq \int_{\Omega}|f|\,d\mu = \lVert f \rVert_{L^1(\Omega)}.$$
Hence integration $I : L^1(\Omega, \mu) \to \mathbb{C}$ is a continuous linear functional with $\lVert I \rVert = 1$.
