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Suppose we have a sequence $\{f_n\}$ of $L^1$ functions such that $||f_n||_1 \leq K_1$, then viewing $L^1(\mathbb{R}) \subset \mathcal{M}(\mathbb{R})$ where $\mathcal{M}(\mathbb{R})$ is the space of Radon measure which is isomorphic to the dual space of $C_C(\mathbb{R})$, we can extract a subsequence $\{{f_n}_k\}$which converges to a Radon measure in the $weak^*$ topology on $\mathcal{M}(\mathbb{R})$. Suppose in addition we have

$\int_{\mathbb{R}}f_n=K_2, \forall n \in \mathbb{N}$ and $f_n \rightarrow f$ pointwise almost everywhere in $\mathbb{R}/\{0\} $

then can we say that $\exists C \in \mathbb{R}$ such that $\int_{\mathbb{R}} {f_n}_k\Phi(x)dx=\int_{\mathbb{R}} f(x)\Phi(x)dx+ C\Phi(0)$

i.e the $weak^*$ limit is of the subsequence is of the form $f+C\delta_{0}$

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Note that convergence almost everywhere in $\mathbb R\setminus\{0\}$ is equivalent to convergence almost everywhere in $\mathbb R$ since $\{0\}$ has measure zero.

Now let $f_n = n \chi_{(1,1+1/n)}$. It converges pointwise a.e. to zero, weak star to $\delta_1$.

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  • $\begingroup$ Thanks.. So the limit is not always of the form $f+C \delta_0$, but can be of the form $f+C \delta_{x_{0}}$ where $x_0 \in \mathbb{R}$ as well. Is this always true? I mean whenever the sequence satisfies the conditions mentioned in my question can we say that the $weak^*$ limit is always of the form $f+C \delta_{x_{0}}$ $\endgroup$ – Rosy Feb 13 at 15:58
  • $\begingroup$ Suppose in addition we have $f_n \rightarrow f$ in $L^1(\mathbb{R}/B(0,\epsilon)) $ for all $\epsilon>0$ then can we say that $\exists C \in \mathbb{R}$ such that $\int_{\mathbb{R}} {f_n}_k\Phi(x)dx \rightarrow \int_{\mathbb{R}} f(x)\Phi(x)dx+ C\Phi(0)$ i.e the $weak^*$ limit in $\mathbb{R}$ of the subsequence is of the form $f+C\delta_{0}$ $\endgroup$ – Rosy Feb 13 at 16:21

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